analysis--complex-analysis-math185.tex

\renewcommand{\bar}{\overline}
\renewcommand{\ddx}{\frac{\partial}{\partial x}}
\newcommand{\dfdx}{\frac{\partial f}{\partial x}}
\newcommand{\dfdy}{\frac{\partial f}{\partial y}}
\newcommand{\dfdzbar}{\frac{\partial f}{\partial \bar z}}
\newcommand{\dfdz}{\frac{\partial f}{\partial z}}
\renewcommand{\dudx}{\frac{\partial u}{\partial x}}
\newcommand{\dudy}{\frac{\partial u}{\partial y}}
\newcommand{\dudz}{\frac{\partial u}{\partial z}}
\renewcommand{\dvdx}{\frac{\partial v}{\partial x}}
\newcommand{\dvdy}{\frac{\partial v}{\partial y}}
\newcommand{\dveczdx}{\frac{\partial \vec z}{\partial x}}
\newcommand{\dveczdy}{\frac{\partial \vec z}{\partial y}}
\newcommand{\dydzbar}{\frac{\partial y}{\partial \bar z}}
\newcommand{\dydz}{\frac{\partial y}{\partial z}}
\newcommand{\dzbar}{\frac{\partial}{\partial \bar z}}
\renewcommand{\ddy}{\frac{\partial}{\partial y}}
\newcommand{\ddz}{\frac{\partial}{\partial z}}
\newcommand{\dxdzbar}{\frac{\partial x}{\partial \bar z}}
\newcommand{\dxdz}{\frac{\partial x}{\partial z}}
\newcommand{\dddfzbarz}{\frac{\partial^2 f}{\partial \bar z \partial z}}
\newcommand{\dddxx}{\frac{\partial^2}{\partial x^2}}
\newcommand{\dddyy}{\frac{\partial^2}{\partial y^2}}
\newcommand{\dddzbarz}{\frac{\partial^2}{\partial \bar z \partial z}}
\newcommand{\ddfdxx}{\frac{\partial^2 f}{\partial x}}
\newcommand{\ddfdyy}{\frac{\partial^2 f}{\partial y}}



% - Berkeley Math 185 - Complex Function Theory (Sarason)

\section*{Useful results}
\textbf{Geometric series} $a + aw + \ldots + aw^n = \frac{a(1 - w^{n+1})}{1-w}$

\section{Complex Numbers}

\begin{description}

% \begin{comment}

\exercise{I.2.1}{Prove that $\C$ obeys the associative law for multiplication
    and the distributive law.}

  Let $u, v, w \in \C$ with $u = a + bi$, $v = c + di$, and $w = f + gi$.

  Multiplication is associative since
  \begin{align*}
    uv &= (ac - bd) - (ad + bc)i \\
       &= (ca - db) - (cb + da)i = vu. \\
  \end{align*}

  Multiplication is left-distributive over addition since
  \begin{align*}
    u(v + w)
    &= (a + bi)\big((c + f) + (d + g)i\big) \\
    &= (ac + af - bd - bg) + (ad + ag + bc + bf)i \\
    &= (ac - bd) + (ad + bc)i + (af - bg) + (ag + bf)i \\
    &= (a + bi)(c + di) + (a + bi)(f + gi) \\
    &=uv + uw.
  \end{align*}

  Since multiplication is commutative, multiplication is also right-distributive over addition.

\exercise{I.2.2}{Find the multiplicative inverses of the complex numbers (0, 1) and (1, 1)}

\exercise{I.2.3}{Think of $\C$ as a vector space over $\R$. Let $c = (a,b)$ be
  in $\C$, and regard multiplication by $c$ as a real linear transformation
  $T_c$. Find the matrix $M_c$ for $T_c$ with respect to the basis
  $(1, 0), (0, 1)$. Observe that the map $c \mapsto M_c$ preserves addition and
  multiplication. Conclude that the algebra of two-by-two matrices over $\R$
  contains a replica of $\C$.}

  Background:

  What does ``$\C$ as a vector space over $\R$'' mean? A vector space is a set of
  tuples. The elements of the tuples are elements of the field ($\R$ in this
  case). Vector spaces support addition and scalar multiplication, where the
  scalars come from the field. So this means that $\C$ is a set of ordered pairs
  of reals, supporting addition of pairs and multiplication of a pair by a real
  scalar. What it does \textit{not} imply is that pairs can be multiplied,
  although, in the case of $\C$, they can, since $\C$ is a field.

  A linear transformation is a function from one vector space $U$ to another,
  $W$, such that $f(u + w) = f(u) + f(w)$, and $f(au) = af(u)$ for $u \in U$,
  $w \in W$ and $a$ in the field. In other words, the linear transformation
  preserves the two vector space operations, addition and scalar multiplication;
  it is a homomorphism on the vector space.

  OK, so the operation that was ignored by conceiving of $\C$ as a vector space
  over $\R$, multiplication of the vectors, we're going to regard as a ``real
  linear transformation'', i.e. a function of $\R^2$. Find the matrix for it with
  respect to the basis $((1, 0), (0, 1))$. The first basis vector ($1$) is
  transformed as $(1, 0) \mapsto (a, b)(1, 0) = (a, b)$. The second basis vector
  ($i$) is transformed as $(0, 1) \mapsto (a, b)(0, 1) = (-b, a)$. Therefore the
  matrix of $T_c$ is
  $$M_c = \matMMxNN{a}{-b}
              {b}{a},$$
  which is a rotation + scaling transformation of $\R^2$.

  \textbf{Observe that the map $c \mapsto M_c$ preserves addition and
    multiplication.}

  Let $f: \R^2 \rightarrow \text{(2x2 matrices)}$ denote the map $c \mapsto M_c$. Then
  $$f(c_1 + c_2) = \matMMxNN{a_1 + a_2}{-b_1 - b_2}
                       {b_1 + b_2}{a_1 + a_2} = f(c_1) + f(c_2),$$
  and
  $$f(c_1c_2) = \matMMxNN{a_1b_1 - a_2b_2}{-a_1b_2 - a_2b_1}
                    {a_1b_2 + a_2b_1}{a_1b_1 - a_2b_2},$$
  while
  $$f(c_1)f(c_2)
  = \matMMxNN{a_1}{-b_1}
        {b_1}{a_1} \matMMxNN{a_2}{-b_2}
                       {b_2}{a_2}
  = \matMMxNN{a_1a_2 - b_1b_2}{-a_1b_2 - a_2b_1}
        {a_2b_1 + a_1b_2}{a_1a_2-b_1b_2}
  $$

  ? That's not preserving multiplication of complex vectors. Does it mean
  preserving scalar multiplication?

  \textbf{Conclude that the algebra of two-by-two matrices over $\R$
    contains a replica of $\C$}

  The ``algebra of two-by-two matrices over $\R$'' refers to the fact that $2x2$
  matrices can be added, and multiplied by a scalar from the field $\R$ (they
  form a vector space), and can also be multiplied.

% \end{comment}

  \exercise{I.4.3}{Prove that if a polynomial with real coefficients has the
    complex root $z$, then it also has $\bar z$ as a root.}

  Let $P:\C \rightarrow \C$ defined by $P(c) = r_0 + r_1c^1 + \ldots + r_kc^k$ be a $k$-th degree
  polynomial of a complex variable $c$, with real coefficients $r_k$, and let
  $z = a + bi$ be a root, i.e. $P(z) = 0$. The claim is that
  $P(\bar z) = 0$.

  To show this, take the complex conjugate of both sides of the equation
  $P(z) = 0$:

  $$
  \bar{P(z)} = \bar{r_0 + r_1z^1 + \ldots + r_kz^k} = \bar 0.
  $$

  Then, since $\bar{z_1 + z_2} = \bar z_1 + \bar z_2$ and $\bar{rz} = r\bar{z}$,

  $$
  r_0 + r_1\bar{z^1} + \ldots + r_k\bar{z^k} = P(\bar z) = 0,
  $$

  proving that if $z$ is a root then $\bar z$ is a root also.

  \exercise{I.7.4}{Prove that the distinct complex numbers $z_1, z_2, z_3$ are
    the vertices of an equilateral triangle if and only if
    $$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1$$}

  The condition can be rewritten as
  $$
  (z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2 = 0,
  $$
  Each of the three terms on the left side is the square of a complex number
  which, when viewed as a vector in $\R^2$, forms one side of a triangle.

  \includegraphics[width=100pt]{img/equilateral-1.png}

  So the original claim is equivalent to the claim that the three vectors
  $$
  (z_1 - z_2)^2, ~ (z_2 - z_3)^2, ~ (z_3 - z_1)^2
  $$
  sum to 0 if and only if the triangle is equilateral.

  First let's prove that if the triangle is equilateral, then the three vectors
  sum to 0. Translate each of the unsquared vectors
  $$
  (z_1 - z_2), ~ (z_2 - z_3), ~ (z_3 - z_1)
  $$
  so that they originate at the origin; they are of equal magnitude and they
  divide the circle into 3 sectors of equal angle $\frac{2\pi}{3}$. Let
  $\theta < \frac{2\pi}{3}$ be the arbitrary angle between one of the vectors and
  the first coordinate axis. Interpreted as complex numbers, we see that their
  arguments are $\theta$, $\frac{2\pi}{3} + \theta$, and
  $\frac{4\pi}{3} + \theta$.

  \includegraphics[width=100pt]{img/equilateral-2.png}

  Now we form their squares
  $$
  (z_1 - z_2)^2, ~ (z_2 - z_3)^2, ~ (z_3 - z_1)^2.
  $$
  Since $(z_1 - z_2)$, $(z_2 - z_3)$, and $(z_3 - z_1)$ are of equal magnitude,
  so are their squares. And the arguments of their squares are $2\theta$,
  $\frac{4\pi}{3} + 2\theta$, and
  $\frac{8\pi}{3} + 2\theta \equiv \frac{2\pi}{3} + 2\theta$ mod
  $2\pi$. Therefore the three squared side vectors, when translated so that
  they originate at the origin, also divide up the circle into sectors of equal
  angle $\frac{2\pi}{3}$: the geometrical picture differs from the previous one
  only by a uniform scaling and relabeling of the vectors, and we conclude that
  these squared vectors also sum to zero (return to the origin when placed
  head-to-tail). I.e. the equilaterality assumption implies
  $$
  (z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2 = 0,
  $$
  proving one direction of the equivalence.

  To prove the other direction, we need to show that if
  $$
  z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1,
  $$
  or equivalently,
  $$
  (z_1 - z_2)^2 + (z_2 - z_3)^2 + (z_3 - z_1)^2 = 0,
  $$
  then the triangle is equilateral. For example, it would suffice to show that
  $$
  |z_1 - z_2| = |z_2 - z_3| = |z_3 - z_1|,
  $$
  but I haven't found a way to do so.

  % The left side of this equation obeys the following triangle inequality
  % $$
  % |z_1^2 + z_2^2 + z_3^2|
  % \le |z_1|^2 + |z_2|^2 + |z_3|^2,
  % $$
  % and the right side obeys
  % $$
  % |z_1z_2 + z_2z_3 + z_3z_1|
  % \le |z_1||z_2| + |z_2||z_3| + |z_3||z_1|.
  % $$

  \exercise{I.10.1}{Use de Moivre's formula to find expressions for
    $\cos 5\theta$ and $\sin 5\theta$ as polynomials in $\cos \theta$ and
    $\sin \theta$.}

  From de Moivre's formula we have
  $(\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta$. The left hand
  side expands as
  \begin{align*}
  (\cos\theta + i\sin\theta)^5 =& \cos^5\theta \\
                               &+ 5i\cos^4\theta\sin\theta \\
                               &- 10\cos^3\theta\sin^2\theta \\
                               &- 10i\cos^2\theta\sin^3\theta \\
                               &+ 5\cos\theta\sin^4\theta \\
                               &+ i\sin^5\theta.
  \end{align*}
  Equating real and imaginary components from the right side and the expansion of
  the left side we have
  \begin{align*}
  \cos 5\theta &= \cos^5\theta - 10\cos^3\theta\sin^2\theta + 5\cos\theta\sin^4\theta \\
  \sin 5\theta &= \sin^5\theta - 10\cos^2\theta\sin^3\theta + 5\cos^4\theta\sin\theta
  \end{align*}
  We can write these as polynomials in $\cos\theta$ and $\sin\theta$
  respectively by using the identity $\cos^2\theta = 1 - \sin^2\theta$:
  \begin{align*}
  \cos 5\theta
  &= \cos^5\theta - 10\cos^3\theta(1 - \cos^2\theta) + 5\cos\theta(1 - 2\cos^2\theta + \cos^4\theta) \\
  &= 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta, \\
  \sin 5\theta
  &= \sin^5\theta - 10(1 - \sin^2\theta)\sin^3\theta + 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta \\
  &= 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta.
  \end{align*}

  \exercise{I.11.4}{Prove that the sum of the $n$-th roots of 1 equals 0,
    ($n > 1$).}

  Let $w = \cos \frac{2\pi}{n} + i\sin \frac{2\pi}{n}$ be the $n$-th root of 1
  with smallest argument, other than 1 itself. Then the sum of the roots is
  $1 + w + w^2 + \ldots + w^{n-1}$. This is the first $n$ terms of a geometric
  series with constant ratio $w$, and is therefore equal to
  $\frac{1 - w^{n}}{1 - w} = \frac{1 - 1}{1 - w} = 0$.


  \exercise{I.11.5}{Let $w$ be an $n$-th root of 1 different from 1
    itself. Establish the formulas
    $$
    1 + 2w + 3w^2 + \ldots + nw^{n-1} = \frac{n}{w -1},
    $$
    $$
    1 + 4w + 9w^2 + \ldots + n^2w^{n-1} = \frac{n^2}{w-1} - \frac{2n}{(w-1)^2}.
    $$
  }

  [Note: my answers to this question appear to be wrong.]

  The sum of the first $n+1$ terms of a geometric series with first term $1$ and
  constant ratio $w$ is

  $$
  1 + w + w^2 + \ldots + w^{n} = \frac{~~~~~1 - w^{n+1}}{1 - w} = \frac{1 - w}{1 - w} = 1,
  $$
  since $w^{n+1} = w$.

  \textcolor{red}{This equation is true for $w$ in $n$-th root, but not for any $w$.}

  Taking derivatives of both sides gives
  $$
  1 + 2w + 3w^2 + \ldots + nw^{n-1} = 0,
  $$

  which does not agree with the given formula, so something's wrong.

  \textcolor{red}{To take derivatives of both sides need to write a function
    identity, not an identity between two numbers.}

  Nevertheless, if it were the case that
  $$
  1 + 2w + 3w^2 + \ldots + nw^{n-1} = \frac{n}{w -1}
  $$
  then we could multiply by $w$, giving
  $$
  w + 2w^2 + 3w^3 + \ldots + nw^n = \frac{nw}{w -1},
  $$
  and differentiate with respect to $w$ again, giving
  $$
  1 + 4w + 9w^2 + \ldots + n^2w^{n-1} = \frac{(w-1)n - nw}{(w - 1)^2} = \frac{n}{w - 1} - \frac{nw}{(w - 1)^2},
  $$
  which also doesn't agree with the given formula.

  \textcolor{red}{Again, to take a derivative, need a function on RHS that
    agrees with LHS $\forall w$.}

% \begin{comment}

\exercise{I.13.1}{Stereographic projection}

  The stereographic projection maps $z$ onto the surface of a sphere according to
  $$
  z \mapsto \frac{(2 \Re z, 2 \Im z, |z|^2 - 1)}{|z|^2 + 1}.
  $$
  % It's 1-1 and the inverse map is
  % $$
  % (\xi, \eta, \zeta) \mapsto (s + 1)\frac{\xi}{2} + (s + 1)\frac{\eta}{2}i
  % $$

\exercise{I.13.1}{Establish the following formula for the spherical metric
$$
\rho(z_1, z_2) = \frac{2|z_1 - z_2|}{\sqrt{|z_1|^2 + 1}\sqrt{|z_2|^2 + 1}}
$$}

  $\rho(z_1, z_2)$
  is the Euclidean distance between the image points of $z_1$ and $z_2$ on the
  Riemann sphere, therefore
  \begin{align*}
  \rho(z_1, z_2)
  &=
  \left|
  \frac{(2 \Re z_1, 2 \Im z_1, |z_1|^2 - 1)}{|z_1|^2 + 1} -
  \frac{(2 \Re z_2, 2 \Im z_2, |z_2|^2 - 1)}{|z_2|^2 + 1}
  \right| \\
  &=
  \left|
  \frac{
  (2 \Re z_1, 2 \Im z_1, |z_1|^2 - 1)(|z_2|^2 + 1) -
  (2 \Re z_2, 2 \Im z_2, |z_2|^2 - 1)(|z_1|^2 + 1)
  }
  {(|z_1|^2 + 1)(|z_2|^2 + 1)}
  \right| \\
  \end{align*}
  Meanwhile,
  $$
  |z_1 - z_2| = \sqrt{(\Re z_1 - \Re z_2)^2 + (\Im z_1 - \Im z_2)^2}
  $$

% \end{comment}

  \exercise{I.14.1}{Establish the formula
    $$
    \rho(z, \infty) = \frac{2}{\sqrt{|z|^2 + 1}}
    $$
  }

  $\rho(z, \infty)$ is the Euclidean distance between the image point of $z$ and
  the north pole $(0, 0, 1)$:
  \begin{align*}
  \rho(z, \infty)
  &= \sqrt{
  \(\frac{2 \Re z}{|z|^2 + 1} - 0\)^2 +
  \(\frac{2 \Im z}{|z|^2 + 1} - 0\)^2 +
  \(\frac{|z|^2 - 1}{|z|^2 + 1} - 1\)^2
  } \\
  &= \frac
  {\sqrt{4 \(\Re z\)^2 + 4 \(\Im z\)^2 + 4}}
  {|z|^2 + 1} \\
  &= \frac
  {2}
  {\sqrt{|z|^2 + 1}}.
  \end{align*}

\end{description}

\section{Complex Differentiation}

Consider $z$ approaching $z_0$. $z - z_0$ is a vector pointing from $z_0$ to
$z$, and $f(z) - f(z_0)$ is a vector pointing between the image points for some
complex-valued function $f$. The derivative of $f$ at $z_0$ is the rotation +
scaling linear transformation (i.e. the complex number $c$) that takes
$z - z_0$ as close as possible to $f(z) - f(z_0)$. Note that the transformation
must be the \textit{same} regardless of the path taken by $z$ as it approaches
$z_0$. In other words, the action of $f$ on \textit{all} vectors in an
infinitisimal disc around $z_0$ is the same as multiplying by a complex number
$c$.

The transformation $f$ can be described by two surfaces over the complex plane:
$u(x,y)$ and $v(x,y)$, so that $f: \cvec{x}{y} \mapsto
\cvec{u(x,y)}{v(x,y)}$. If $f$ is differentiable at $(x_0, y_0)$ then it has a
local linear approximation with sublinear error. That linear approximation is
$$
\cvec{u(x, y)}
{v(x, y)} \approx \cvec{u(x_0, y_0)}
{v(x_0, y_0)} + \cvec{(x - x_0)\dudx + (y - y_0)\dudy}
{(x - x_0)\dvdx + (y - y_0)\dvdy}
$$
This is more succinctly expressed using the Jacobian:
$$
\cvec{u(x, y)}
{v(x, y)} \approx \cvec{u(x_0, y_0)}
{v(x_0, y_0)} + \matMMxNN{u_x}{u_y}
{v_x}{v_y} \cvec{x - x_0}
{y - y_0}.
$$
Note that this "linear approximation" form
$$
y \approx y_0 + y'(x - x_0)
$$
could just as well be written
$$
y - y_0 \approx y'(x - x_0)
$$
showing that one way of describing the derivative is "whatever you have to
multiply a small displacement in the input space by to get the displacement in
the output space".

Recall that the derivative of a complex function $f$ is defined to be a complex
number,
$$
f'\(\cvec{x_0}
{y_0}\) = \lim_{(x,y) \rightarrow (x_0,y_0)}
\frac{
  \cvec{u(x, y)}
  {v(x, y)} - \cvec{u(x_0)}
  {v(y_0)}
}
{
  \cvec{x}
  {y} - \cvec{x_0}
  {y_o}.
},
$$
i.e.
$$
f'(z_0) = \lim_{z \rightarrow z_0} \frac{f(z) - f(z_0)}{z - z_0},
$$
i.e. the derivative is whatever complex number you multiply the vector $z-z_0$
by to get its image vector $f(z) - f(z_0)$, in the limit as $z \rightarrow z_0$.

The partial derivatives of the complex-valued $f$ in the real and imaginary
directions are the complex numbers
\begin{align*}
f_x &= u_x + iv_x\\
f_y &= u_y + iv_y\\
\end{align*}
or
\begin{align*}
f_x &= \cvec{u_x}{v_x}\\
f_y &= \cvec{u_y}{v_y}\\
\end{align*}

The geometric interpretation of these is that they define how the image vector
$f(z)$ changes in response to a small change to $z$.

$u$ can be approximated by a local tangent plane. That's what $u_x$ and $u_y$
do. And so can $v$; that's what $v_x$ and $v_y$ do. But when we consider the
effect of a small displacement in the 2D input space on the 2D output space, we
describe the two tangent plane approximations jointly as a linear
transformation of the input plane, defined by the Jacobian. The thing is, the
linear transformation must have the same effect as multiplication by a complex
number.

The derivative is "what you have to multiply the input displacement by to get
the output displacement". That's true for a single-variable function
$\R \rightarrow \R$
$$
u(x) - u(x_0) = f'(x_0) \cdot (x - x_0)
$$
and it's true for a surface over the plane $(\R^2 \rightarrow \R$)
$$
u(x, y) - u(x_0, y_0) = \dudx \cdot (x - x_0) + \dudy \cdot (y - y_0)
$$
so presumably something analogous holds for a linear transformation of the
plane $(\R^2 \rightarrow \R^2$), i.e.
$$
\vec z - \vec z_0 = \dveczdx \cdot (x - x_0) + \dveczdy \cdot (y - y_0).
$$
or
$$
\cvec{u(x, y)}{v(x, y)} - \cvec{u(x_0, y_o)}{v(x_0, y_0)} =
\cvec{u_x}{v_x} \cdot (x - x_0) + \cvec{u_y}{v_y} \cdot (y - y_0).
$$
That's exactly the same as the equation involving the Jacobian above
$$
\cvec{u(x, y)}
{v(x, y)} - \cvec{u(x_0, y_0)}
{v(x_0, y_0)} = \matMMxNN{u_x}{u_y}
{v_x}{v_y} \cvec{x - x_0}
{y - y_0}.
$$

So how are we to make sense of the equation relating $f'$ and the partial
derivatives $\dfdx$ and $\dfdy$? Clearly in some sense the Jacobian \textit{is}
$f'$, or at least, the complex number that does what the Jacobian does is
$f'$. And
\begin{align*}
\dfdx &= \cvec{u_x}
              {v_x} = u_x + iv_x \\
\dfdy &= \cvec{u_y}
              {v_y} = u_y + iv_y,
\end{align*}
and so from the Cauchy-Riemann constraint
\begin{align*}
\dfdy &= -v_x + iu_x = i\dfdx,
\end{align*}
i.e. the partial derivative w.r.t. $y$ points at 90° to the $x$ partial
derivative.

% $f_x$ is the complex number that transforms a small real displacement vector
% $\epsilon$ to its image, and $f_y$ is the complex number that transforms a
% small imaginary displacement vector $i\delta$ to its image. [But shouldn't
% these be the same and equal to $f'$, seeing as $f'$ rotates and scales all
% displacement vectors in an infinitesimal disc uniformly? Instead, C-R says that
% $f_x = -if_y$.]

% One notation for the "directional derivative" of $f$ in the direction of
% $\vec z$ is $\frac{\partial f}{\partial \vec z}$. Is this an entirely separate
% notion from the partial derivatives $\frac{\partial f}{\partial z}$ and
% $\frac{\partial f}{\partial \bar z}$ considered in the context of complex
% analysis / Wirtinger derivatives?

So if the local linear approximation to the transformation $f$ behaves exactly as
multiplication by a complex number, then the Jacobian must have the form of a
rotation+scale matrix, $\smat{a}{-b}
{b}{a}$. Therefore the Jacobian must satisfy the
Cauchy-Riemann equations
$$
\begin{cases}
  u_x = v_y \\
  v_x = -u_y. \\
\end{cases}
$$
The Jacobian that effects the local linear rotation+scale transformation, together with
the equivalent complex number, is
$$
\matMMxNN{u_x}{-v_x}
{v_x}{u_x} ~~~~ u_x + iv_x
$$
or
$$
\matMMxNN{v_y}{u_y}
{-u_y}{v_y} ~~~~ v_y - iu_y.
$$
So we can write
\begin{align*}
f'
&= u_x + iv_x = f_x \\
&= v_y - iu_y = -if_y,
\end{align*}
therefore as above, another expression of the Cauchy-Riemann criterion is
$$
f_x = -if_y.
$$
Question: what is the intuition for the fact that the complex number
representing the partial derivative with respect to $x$ is the \textit{same} as
the complex number that effects the full linear transformation? (and at 90° to
the partial with respect to $y$) And what's the intuition for the fact that
$\dfdz = \dfdx$, while $\dfdzbar = 0$?

% The partial derivatives of the complex-valued $f$ in the real and imaginary
% directions are
% \begin{align*}
% \dfdx &= \dudx + i\dvdx\\
% \dfdy &= \dudy + i\dvdy\\
% \end{align*}

% So we can write
% \begin{align*}
% f'(z_0)
% &= \dudx(z_0) + i \dvdx(z_0) =   \dfdx(z_0) \\
% &= \dvdy(z_0) - i \dudy(z_0) = -i\dfdy(z_0),
% \end{align*}
% or
% \begin{align*}
% f'\(\cvec{x_0}{y_0}\)
% &= \cvec{\dudx(z_0)}{\dvdx(z_0)}  =   \dfdx(z_0)  \\\\
% &= \cvec{\dvdy(z_0)}{-\dudy(z_0)} = -i\dfdy(z_0).
% \end{align*}

$f$ is differentiable iff the error in the linear transformation goes to $0$ as
$(x,y) \rightarrow (x_0,y_0)$ (i.e. real partial derivatives of $u$ and $v$
exist) and the partial derivatives satisfy the Cauchy-Riemann equations.

\subsection*{Partial derivatives in the $z$ and $\bar z$ directions}
The (fixed) $x, y$ and (varying) $z, \bar z$ directions are related by
\begin{align*}
x &= (z + \bar z)/2 \\
y &= (z - \bar z)/2i.
\end{align*}
So by the chain rule,
\begin{align*}
\dfdz
&= \dfdx \dxdz + \dfdy \dydz \\
&= (u_x + iv_x)\frac{1}{2} + i(u_x + iv_x)\frac{1}{2i} \\
&= u_x + iv_x \\
&= \dfdx
\end{align*}
and
\begin{align*}
\dfdzbar
&= \dfdx \dxdzbar + \dfdy \dydzbar \\
&= (u_x + iv_x)\frac{1}{2} + i(u_x + iv_x)\frac{-1}{2i} \\
&= \frac{1}{2}\Big((u_x - u_x) + i(v_x - v_x)\Big) \\
&= 0.
\end{align*}

\begin{description}

  \exercise{II.8.1(b,d)}{
    Let the function $f$ be holomorphic in the open disc $D$. Prove that each of
    the following conditions forces $f$ to be constant:\\\\
    \textnormal{Let $f(z) = u(z) + iv(z)$.}
    \begin{description}
      \exercise{(a)}{$f' = 0$ throughout $D$} \\\\
      \textnormal{ \textbfit{Informally:} $f' = 0$ throughout $D$ means that the
        best linear approximation of $f(z) - f(z_0)$ is $0(z - z_0)$ which
        implies that $f(z) = f(z_0)$ everywhere, so $f$ is constant.\\
        ~\\
        \textbfit{Formally:} Since $f$ is holomorphic, $f' = u_x + iv_x =
        0$. Equating real and imaginary parts shows that $u_x = v_x = 0$ and
        therefore that $v_y = u_x = 0$ and $u_y = -v_x = 0$. Since the Jacobian
        of $f$ is the zero matrix, $f$ is constant.}\\
      \exercise{(b)}{$f$ is real-valued in $D$} \\\\
      \textnormal{
        % \textbfit{Informally:}
        $f$ is real-valued, so $f(z) - f(z_0)$ is real-valued. Therefore the
        local linear approximation $c(z - z_0)$ collapses the plane onto the real
        axis, i.e. the Jacobian matrix has the form $\matMMxNN{a}{b} {0}{0}$. But $f$
        is holomorphic, so the Jacobian must also have the form
        $\matMMxNN{a}{-b}
        {b}{a}$. Therefore the Jacobian is the zero matrix, i.e.
        all partial derivatives are zero, $u_x = u_y = v_x = v_y = 0$, so $f$ is constant.\\
        % \textbfit{Formally:} $f$ is real-valued means that $v(z) = 0$
        % everywhere. Therefore the partial derivatives $v_x = v_y = 0$. Since $f$
        % is holomorphic the partial derivatives satisfy the Cauchy-Riemann
        % equations, therefore we have $u_x = v_y = 0$ and $u_y = -v_x = 0$. The
        % Jacobian is the zero matrix hence $f$ is constant.
        % \\
      }
      \exercise{(c)}{$|f|$ is constant in $D$} \\\\
      \textnormal{ \textbfit{Informally:} $|f|$ is constant means that it collapses
        all points in the open disc $D$ onto a circle. Therefore the Jacobian of
        $f$ has determinant $0 = u_x^2 + v_x^2 = v_y^2 + u_y^2$. Therefore the
        Jacobian
        is the zero matrix and the function $f$ is constant.\\
        ~\\
        \textbfit{Formally:} $|f|$ is constant, therefore
        $|f|^2 = f\bar f = u^2 + v^2$ is constant. Therefore the following two
        partial derivatives are constant:
        \begin{align*}
        \begin{cases}
          \ddx |f|^2 =  2uu_x + 2vv_x = 0 \\
          \ddy |f|^2 = 2uu_y + 2vv_y = 0.
        \end{cases}
      \end{align*}
      Since $f$ is holomorphic, $u_x = v_y$ and $u_y = -v_x$, so
      \begin{align*}
        \begin{cases}
          uu_x - vu_y = 0 \\
          uu_y + vu_x = 0,
        \end{cases}
      \end{align*}
      % i.e.
      % $$
      % \matMMxNN{u}{-v}
      % {u}{v} \cvec{u_x}{u_y} = \cvec{0}{0}.
      % $$
      % The determinant of this system is $2uv$ which is not in general zero. But
      % the equations hold throughout the disc $D$, therefore
      Multiplying the first equation by $u$ and the second by $v$ we have
      $$
      \begin{cases}
        u^2u_x - uvu_y = 0 \\
        uvu_y + v^2u_x = 0,
      \end{cases}
      $$
      and summing these gives
      $$
      u_x(u^2 + v^2) = 0,
      $$
      which proves that either $u_x = 0$ or that $f$ is constant (in which case $u_x = 0$ also).\\
      Similarly, multiplying the first equation by $v$ and the second by $u$ gives
      $$
      \begin{cases}
        uvu_x - v^2u_y = 0 \\
        u^2u_y + uvu_x = 0,
      \end{cases}
      $$
      and subtracting the first from the second gives
      $$
      u_y(u^2 + v^2) = 0.
      $$
      We conclude that $u_x = u_y = 0$ and that $f$ is therefore constant.  }
    \exercise{(d)}{$\arg f$ is constant in $D$} \\\\
    \textnormal{
      % \textbfit{Informally:}
      Let $\arg f = \theta$, constant throughout
      $D$. Then $\arg (f(z) - f(z_0)) = \theta$, whenever $z \neq z_0$.
      Therefore the best local linear approximation to $f$ is a linear
      transformation that collapses the plane onto a line with angle $\theta$.
      The Jacobian determinant is therefore zero. Since $f$
      is holomorphic the Jacobian is of the form $\matMMxNN{a}{-b}
      {b}{a}$ and therefore we have
      $a^2 + b^2 = 0$, so $a = b = 0$.  Therefore the Jacobian is the zero
      matrix, i.e. $f' = 0$ throughout $D$, so $f$ is constant.\\
      % ~\\
      % \textbfit{Formally:}
      % The polar form of the Cauchy-Riemann equations is
      % \begin{align*}
      % v_\theta &= ru_r \\
      % u_\theta &= -rv_r.
      % \end{align*}
      % Since $\arg f$ is constant, $v_\theta$
    }
  \end{description}
}

\exercise{II.8.2}{
  Let the function $f$ be holomorphic in the open set $G$. Prove that the
  function $g(z) = \bar{f(\bar z)}$ is holomorphic in the set
  $G^* = \{\bar z: z \in G\}$.
} \\\\
Let
\begin{align*}
f: x + iy &\mapsto s(x, y) + i t(x,y). \\
g: x + iy &\mapsto u(x, y) + i v(x,y) \\
\end{align*}
We want to show that the Jacobian of $g$ exists and and satisfies the
Cauchy-Riemann equations. We have
\begin{align*}
g(x + iy)
&= \bar{s(x, -y) + i t(x, -y)} \\
&= s(x, -y) - i t(x, -y),
\end{align*}
and therefore
\begin{align*}
  u(x,y) &= s(x, -y) \\
  v(x, y) &= -t(x, -y).
\end{align*}
Now $f = s + it$ is holomorphic, so $s_x=t_y$ and $s_y = -t_x$. Therefore the
partial derivatives of $g$ are
\begin{align*}
  u_x &= \ddx s(x, -y)  = s_x \\
  u_y &= \ddy s(x, -y)  = -s_y = t_x \\
  v_x &= -\ddx t(x, -y) = -t_x \\
  v_y &= -\ddy t(x, -y) = t_y = s_x. \\
\end{align*}
Therefore $u_x = v_y$ and $v_x = -u_y$, showing that the Jacobian of $g$
satisfies the Cauchy-Riemann equations, and therefore that $g$ is holomorphic
in its domain.

\exercise{II.16.4}{
  Prove that, if $u$ is a real-valued harmonic function in an open disk $D$,
  then any two harmonic conjugates of $u$ in $D$ differ by a constant.
}

Let $v$ and $w$ be harmonic conjugates of $u$, so that
$$
\begin{cases}
  u_x = v_y = w_y\\
  u_y = -v_x = -w_x.
\end{cases}
$$
We want to show that $q = v-w$ is constant, i.e. that $q_x = q_y = 0$,
throughout $D$. From the Cauchy-Riemann equalities above, we have
$q_x = v_x - w_x = 0$ and $q_y = v_y - w_y = 0$ as required.

% \exercise{II.16.5}{
%   [Not in homework.] Suppose that $u$ is a real-valued harmonic function in an open disk $D$, and
%   suppose that $u^2$ is also harmonic. Prove that $u$ is constant.
% }

\exercise{II.16.7}{
  Prove (assuming equality of second-order mixed partial derivatives) that
  $$
  \dddzbarz =
  \frac{1}{4}\(\frac{\partial^2}{\partial x^2} +
  \frac{\partial^2}{\partial y^2}\)
  $$
  Thus, Laplace's equation can be written as $\frac{\partial^2 f}{\partial \bar z \partial z} = 0$.
}

First note that $x$ and $y$ are related to $\bar z$ via
\begin{align*}
x = \frac{z + \bar z}{2} \\
y = \frac{z - \bar z}{2i}, \\
\end{align*}
therefore by the chain rule
\begin{align*}
\dzbar
&= \ddx \dxdzbar = \frac{1}{2}\ddx \\
&= \ddy \dydzbar = -\frac{1}{2i}\ddy. \\
\end{align*}

Now $\ddz$ is defined by
$$
\ddz = \frac{1}{2}\(\ddx - i\ddy\),
$$
and taking the partial derivative with respect to $\bar z$ gives
\begin{align*}
\dddzbarz
&= \frac{1}{2}\(\dzbar \ddx - i \dzbar \ddy\) \\
&= \frac{1}{2}\(\frac{1}{2}\ddx \ddx - i \(\frac{-1}{2i}\)\ddy \ddy\) \\
&= \frac{1}{4}\(\dddxx + \dddyy\).
\end{align*}

Laplace's equation is $\ddfdxx + \ddfdyy = 0$, which can also be written as
$$
\(\dddxx + \dddyy\)f = 0,
$$
and therefore
$$
4 \dddzbarz f = 0,
$$
i.e.
$$
\dddfzbarz = 0.
$$

\exercise{II.16.8}{
  Prove that if $u$ is a real-valued harmonic function then the function
  $\frac{\partial u}{\partial z}$ is holomorphic.
}
As above, first note that $x$ and $y$ are related to $\bar z$ via
\begin{align*}
x &= \frac{z + \bar z}{2} \\
y &= \frac{z - \bar z}{2i} = -i\frac{z - \bar z}{2}. \\
\end{align*}
By the chain rule
\begin{align*}
\dudz
&= \dudx \dxdz + \dudy \dydz \\
&= \frac{1}{2} \dudx  - \frac{i}{2} \dudy. \\
\end{align*}
Switching notation, we write this as $\dudz = \frac{1}{2} u_x  - \frac{i}{2} u_y$.

Define a complex-valued function
\begin{align*}
w(x + iy)
= \dudz &= s(x,y) + it(x,y) \\
&= \frac{1}{2} u_x  - \frac{i}{2} u_y.
\end{align*}
Then the Jacobian of $w$ is
\begin{align*}
\matMMxNN{s_x}{s_y}
    {t_x}{t_y} = \frac{1}{2}\matMMxNN{u_{xx}}{u_{xy}}
                                {-u_{yx}}{-u_{yy}}.
\end{align*}
But since $u$ is harmonic, we know that $u_{xx} + u_{yy} = 0$, therefore the
Jacobian of $w$ satisfies the Cauchy-Riemann equations and $w = \dudz$ is
holomorphic.

\end{description}

\section{Image of a curve under a transformation}
What is the effect of the inversion mapping $z \mapsto w = \frac{1}{z}$ on circles
and lines?

Let $z = x + iy$ with image $w = \frac{1}{z} = u + iv$ and note that
$\frac{1}{z} = \frac{\bar z}{|z|^2}$. Therefore the mapping is
\begin{align*}
  x + iy \mapsto \frac{x}{x^2 + y^2} - i\frac{y}{x^2 + y^2} = u + iv.
\end{align*}

The general equation of a circle or line in the plane is
\begin{align*}
  Ax^2 + Ay^2 + Bx + Cy + D  = 0.
\end{align*}

We use the inverse mapping to establish an equation that holds in the
transformed complex plane. Since the inverse mapping is the same as the forward
mapping, we have
\begin{align*}
  w = u + iv \mapsto \frac{u}{u^2 + v^2} - i\frac{v}{u^2 + v^2} = x + iy.
\end{align*}
So points $w = u + iv$ in the transformed complex plane satisfy
\begin{align*}
  A\frac{u^2}{(u^2 + v^2)^2} + A\frac{v^2}{(u^2 + v^2)^2} + B\frac{u}{u^2 + v^2} - C\frac{v}{u^2 + v^2} + D  = 0,
\end{align*}
i.e.
\begin{align*}
  \frac{A}{u^2 + v^2} + B\frac{u}{u^2 + v^2} - C\frac{v}{u^2 + v^2} + D  = 0,
\end{align*}
or
\begin{align*}
  A + Bu - Cv + Du^2 + Dv^2  = 0.
\end{align*}
So we see that, if a circle/line exists in the pre-transformed plane, then...



\section{Linear-Fractional Transformations}

Complex projective space $\CP^1$ is a space of equivalence classes of vectors
in $\C^2$. Basically the elements of $\CP^1$ are analogs of lines through the
origin in $\R^2$ (one-dimensionsal subspaces): two vectors are equivalent if
the ratios between their vector components are equal. And that ratio provides a
bijection between $\CP^1$ and $\bar \C$.

Since linear transformations of $\C^2$ map lines (in $\C^2$) to lines (in
$\C^2$), they induce a bijection on $\CP^1$ and therefore on $\bar \C$.

In fact linear-fractional transformations are induced by a two-by-two complex
matrix (an element of $\GLC{2}$) [Do I understand why?]. This makes
linear-fractional transformations closed under composition and gives them an
identity (the LFT corresponding to the identity matrix) and inverses (given by
the matrix inverse). So there is a group of LFTs which is the homomorphic image
of $\GLC {2}$, under the map which sends a two-by-two matrix to its induced
LFT. The kernel of the homomorphism contains scalar multiples of the identity
matrix $I_2$. I think that's basically because such uniform scaling matrices
leave lines unchanged and therefore leave the one-dimensional subspaces
unchanged. Therefore the group of LFTs is isomorphic to the quotient group
$\GLC{2}/(\C\backslash\{0\})I_2$ (each coset is formed by taking a matrix and
scaling it by multipying it with a scaled identity matrix from the kernel).

\begin{description}
  \exercise{III.5.2}{
    ~\\
    Given four distinct points $z_1, z_2, z_3, z_4$ in $\bar \C$, their cross
    ratio, which is denoted by $(z_1, z_2; z_3, z_4)$ is defined to be the
    image of $z_4$ under the linear-fractional transformation that sends
    $z_1,z_2,z_3$ to $\infty, 0, 1$, respectively. Prove that if $\phi$ is a
    linear-fractional transformation then
    $$\(\phi(z_1),\phi(z_2);\phi(z_3),\phi(z_4)\) = (z_1, z_2; z_3, z_4).$$
    ~\\
  }\\
  Let $f$ be the linear-fractional transformation that maps $z_1, z_2, z_3$ to
  $\infty, 0, 1$ respectively, so that the cross-ratio is defined to be
  $(z_1, z_2; z_3, z_4) = f(z_4)$. We want to show that the cross ratio,
  defined in this way, is invariant under an arbitrary linear-fractional
  transformation $\phi$.

  First, let's find an explicit expression for $f(z)$ in terms of
  $z_1,z_2,z_3$. We know that $f(z_1) = \infty$ and $f(z_2) = 0$, so perhaps
  $f$ has the form $f(z) = c\frac{z_2 - z}{z_1 - z}$ for some constant $c$. We
  also require $f(z_3) = 1$. One way to achieve that is to choose
  $c = \frac{z_1 - z_3}{z_2 - z_3}$, so the definition of $f$ becomes
  $$
  f(z) = c\frac{(z_2 - z)(z_1 - z_3)}{(z_2 - z_3)(z_1 - z)}.
  $$
  Defined like this, $f$ is a linear-fractional transformation, and it does
  send $z_1,z_2,z_3$ to $\infty, 0, 1$, respectively. Furthermore, by theorem
  III.5, this is the only linear-fractional transformation that does so.

  So we have
  $$
  (z_1, z_2; z_3, z_4) = f(z_4) = \frac{(z_1 - z_3)(z_2 - z_4)}{(z_1 - z_4)(z_2 - z_3)},
  $$
  and we want to show that this quantity is invariant under an arbitrary
  linear-fractional transformation $\phi$. Let
  $\phi(z) = \frac{az + b}{cz + d}$, with $ad - bc = 1$ (since we are free to
  scale the coefficients $a,b,c,d$ uniformly as we wish, if $ad - bc \ne 1$
  then we scale them all by $\frac{1}{\sqrt{ad - bc}}$). Now consider
  \begin{align*}
    \phi(z_i) - \phi(z_j)
    &= \frac{(az_i + b)(cz_j + d) - (az_j + b)(cz_i + d)}{(cz_i + d)(cz_j + d)} \\
    &= \frac{z_iz_j(ac - ac) + z_i(ad - bc) + z_j(bc - ad) + (bd - bd)}{(cz_i + d)(cz_j + d)} \\
    &= \frac{z_i - z_j}{(cz_i + d)(cz_j + d)}.
  \end{align*}
  Letting $A_i = cz_i + d$, we see that the cross-ratio of the transformed
  points is
  \begin{align*}
    \(\phi(z_1),\phi(z_2);\phi(z_3),\phi(z_4)\) = \frac{(z_1 - z_3)(z_2 - z_4)/A_1A_3A_2A_4}{(z_1 - z_4)(z_2 - z_3)/A_1A_4A_2A_3} = (z_1, z_2; z_3, z_4).
  \end{align*}

  % $$
  % z_1, z_2, z_3 = -\frac{d}{c}, \frac{b}{a}, \frac{+d - b}{-c + a}
  % $$

  \exercise{III.6.3}{
    ~\\
    Prove that a linear-fractional transformation with only one fixed point is
    conjugate to a translation.
    ~\\
  }\\
  Let $\phi(z) = \frac{az + b}{cz + d}$, with $ad - bc = 1$ (justified in
  III.5.2 above). The fixed points of this mapping are the solutions of
  \begin{align*}
    \frac{az + b}{cz + d} = z,
  \end{align*}
  which is a quadratic equation
  $$
  cz^2 + (d - a)z - b = 0,
  $$
  with solutions
  \begin{align*}
  z &= \frac{(a - d) \pm \sqrt{(a - d)^2 + 4bc}}{2c} \\
    &= \frac{(a - d) \pm \sqrt{(a + d)^2 - 4(ad - bc)}}{2c} \\
    &= \frac{(a - d) \pm \sqrt{(a + d)^2 - 4}}{2c} \\
  \end{align*}
  $\phi_1$ has only one fixed point, so $a + d = \pm 2$, i.e. $d = 2 - a$ or $d = -2 - a$.

  We want to show that there exists a linear-fractional transformation $\phi_2(z) = z + k$,
  and another linear-fractional transformation $\psi$, such that
  $\phi_2 = \psi \circ \phi_1 \circ \psi^{-1}$. In other words, performing the
  $\phi_1$ transformation under the change of basis specified by $\psi$, yields
  a translation.

  Let's just try to show that by calculation. Let
  $\psi(z) = \frac{ez + f}{gz + h}$ with $eh - fg = 1$ so that
  $\psi^{-1}(z) = \frac{hz -f}{-gz + e}$. Then
  \begin{align*}
    \phi_2(z)
    &= (\psi \circ \phi_1 \circ \psi^{-1})(z) \\
    &= \matMMxNN{e}{f}
           {g}{h} \matMMxNN{a}{b}
                      {c}{d} \matMMxNN{h }{-f}
                                 {-g}{e } \cvec{z}{1} \\
    &= \matMMxNN{e}{f}
           {g}{h} \matMMxNN{ ah - bg}{-af + be}
                      {-ch - dg}{-cf + de} \cvec{z}{1} \\
    &= \matMMxNN{e(ah - bg) + f(-ch - dg)}{e(-af + be) + f(-cf + de)}
           {g(ah - bg) + h(-ch - dg)}{g(-af + be) + h(-cf + de)} \cvec{z}{1} \\
  \end{align*}

  Things we know:
  \begin{itemize}
  \item Trace is invariant under change of basis, so the trace of the product
    of the 3 matrices is the same as that of $\phi_1$: $\pm 2$.
  \item The determinants of all the matrices and products thereof are 1.
  \end{itemize}
  So
  \begin{align*}
    e(ah - bg) + f(-ch - dg) + g(-af + be) + h(-cf + de) &= \pm 2 \\
    (ah - bg)(-cf + de) - (-af + be)(-ch - dg) &= 1 \\
    eh - fg &= 1 \\
    ad - bc &= 1 \\
    a + d &= \pm 2
  \end{align*}
  A translation has matrix of the form $\matMMxNN{x}{y}
  {0}{x}$. So the question is,
  can we find $e,f,g,h$ such that
  \begin{align*}
    g(ah - bg) + h(-ch - dg) &= 0 \\
    e(ah - bg) + f(-ch - dg) &= g(-af + be) + h(-cf + de)
  \end{align*}

  \begin{align*}
    ah - bg = h(ch + dg)/g \\
    eh(ch + dg)/g + f(-ch - dg) &= g(-af + be) + h(-cf + de)
  \end{align*}

  % \exercise{III.9.1}{
  % ~\\
  % Find the image of the half-plane $\Re z > 0$ under the linear-fractional
  % transformation that maps $0, i, -i$ to $1, -1, 0$, respectively.
  % ~\\
  % }\\
  %   Let the linear-fractional transformation be $\phi(z) = \frac{az + b}{cz + d}$. We have
  %   \begin{align*}
  %   \phi(-i) &= \frac{-ai + b}{-ci + d} = 0 \implies b = ai \\
  %   \phi(0) &= \frac{b}{d} = 1 \implies d = b = ai \\
  %   \phi(i) &= \frac{ai + b}{ci + d} = -1 \implies ci + d = -2b \implies c = -3a,
  % \end{align*}
  % so
  % \begin{align*}
  %   \phi(z) = \frac{az + ai}{-3az + ai} = \frac{z + i}{-3z + i}.
  % \end{align*}
  % % Now let's find how $z = x + iy$ is transformed:
  % % \begin{align*}
  % %   \phi(x + iy)
  % %   &= \frac{x + iy + i}{-3(x + iy) + i} \\
  % %   &= \frac{x + i(y + 1)}{-3x + i(-3y + 1)} \\
  % %   &= \frac{(x + i(y + 1))(-3x - i(-3y + 1))}{9x^2 + (-3y + 1)^2} \\
  % %   &= \frac{\Big(-3x^2 + (y + 1)(-3y + 1)\Big) + i\Big((y+1)(-3x) - x(-3y + 1)\Big)}{9x^2 + (-3y + 1)^2} \\
  % %   &= \frac{\Big(-3x^2 + (y + 1)(-3y + 1)\Big) - 4xi}{9x^2 + (-3y + 1)^2}.
  % % \end{align*}
  % Points on the line $\Re z = 0$ are transformed as
  % \begin{align*}
  %   \phi(iy) = \frac{iy + i}{-3iy + i} = \frac{y + 1}{-3y + 1},
  % \end{align*}
  % so they are mapped onto the real line for $y \ne \frac{1}{3}$.
  \exercise{III.9.2}{
    ~\\
    Find the images of the disc $|z| < 1$ and the half-plane $\Re z > 0$ under
    the linear-fractional transformation that maps $\infty$ to $1$ and has $i$
    and $-i$ as fixed points.
    ~\\
  }\\
  The boundary of the disc is the unit circle and therefore must be mapped to
  either a circle or a line. If it were mapped to a circle then not only $i$
  and $-i$ would be fixed points but also all points on the unit circle in the
  domain would be fixed and, in particular, $1$ would be a fixed point. But $1$
  is not fixed since $\infty \mapsto 1$. Therefore the unit circle is mapped to
  a line and this line must be the imaginary axis $\Re z = 0$, since it must
  contain $i$ and $-i$.

  The image of the disc $|z| < 1$ must therefore be one of the half-planes
  either side of the imaginary axis, since connected sets are mapped to
  connected sets. To determine which, we note that $\infty$ is mapped to
  $1$. But $\infty$ was outside the unit disc in the domain, and so in the
  transformed complex plane, the image of $\infty$ must remain connected to
  points outside the image of the unit disc. Therefore the image of the unit
  disc is the half plane $\Re z < 0$.

  The imaginary axis $\Re z = 0$ must be mapped to a circle, since we know that
  its image contains $i, -i$ and $1$, and no line passes through those 3
  points. In fact, its image must be the unit circle, i.e. the equator on the
  Riemann sphere. So the remaining question is whether the image of the
  half-plane $\Re z > 0$ is the northern or southern hemisphere. To determine
  which we note that, before the transformation, it was possible to walk North
  from $-i$, through $\infty$, to $i$, with the half-plane in question on our
  right-hand side. Therefore the image of the half-plane must$^1$ be on the
  same side as we perform the corresponding walk between the images of those
  points. That walk takes us from $-i$, through $1$, to $i$, showing that the
  image of the half-plane $\Re z > 0$ is the southern hemisphere $|z|$,
  i.e. the unit disc $|z| < 1$.

  $^1$ This argument is based on a theorem stating that linear-fractional
  transformations are conformal, and the definition of conformality specifying
  that orientations of the sort described are preserved.

  \exercise{III.9.5}{
    ~\\
    Prove that the linear-fractional transformations mapping the disc $|z| < 1$
    onto itself are those induced by matrices of the form
    $$
    \matMMxNN{a     }{b     }
    {\bar b}{\bar a}
    $$
    with $|a|^2 - |b|^2 = 1$.
    ~\\
  }
  \\
  My initial thought here was the following:\\
  \\
  The transformation maps the unit disc (i.e. the southern hemisphere of the
  Riemann sphere) onto itself. In order for that to be so, I suspect it would
  have to map the unit circle onto itself (perhaps an argument based on
  continuity of the transformation here?). And I think that a linear-fractional
  transformation maps the unit circle onto itself if and only if that mapping
  is multiplication by a unit-length complex number i.e. rotation of the
  Riemann sphere around the polar axis. Such mappings have the form
  \begin{align*}
    f(z) = \frac{az + b}{cz + d}
  \end{align*}
  where $b = c = 0$ and $|a| = |d|$. So an answer along those lines would hope to show that a
  linear-fractional transformation is induced by a matrix of the form
  $$
  \matMMxNN{a     }{b     }
  {\bar b}{\bar a}
  $$
  with $|a|^2 - |b|^2 = 1$, if and only if the matrix is of the form
  \begin{align*}
    \matMMxNN{a}{0}
        {0}{d},
  \end{align*}
  with $|a| = |d|$. But that doesn't look to be a true statement.
  \\
  % \begin{align*}
  %   f(z)
  %   &= \frac{az + b}{\bar b z + \bar a} \\
  %   &= \frac{(az + b)()}{\bar b z + \bar a}
  % \end{align*}
  \exercise{III.9.7}{
    ~\\
    For the function
    $$
    f(z) = \(\frac{z + 1}{z - 1}\)^2
    $$
    (defined to equal $1$ at $z = \infty$ and $\infty$ at $z = 1$), find the
    images of the following sets:
    \begin{description}
    \item{(a) The extended real axis.}
    \item{(b) The extended imaginary axis.}
    \item{(c) The half-plane $\Re z > 0$.}
    \end{description}
  }
  ~\\
  \begin{align*}
    f(z)
    &= \(\frac{z + 1}{z - 1}\)^2 \\
    &= \frac{z^2 + 2z + 1}{z^2 - 2z + 1} \\
    &= w
  \end{align*}
  We have
  \begin{align*}
    0 &\mapsto 1 \\
    \infty &\mapsto 1 \\
    1 &\mapsto \infty \\
    -1 &\mapsto 0 \\
    i &\mapsto -1 \\
    -i &\mapsto \frac{i}{i + 2}
  \end{align*}
  so the mapping is non-injective and therefore non-invertible.

  Also as far as I can see the mapping can not be viewed as a linear-fractional
  transformation as these are ratios of first-degree, not second-degree,
  polynomials in $z$. Therefore I can't use any theorems about
  linear-fractional transformations such as triple transitivity and
  preservation of clircles.

  One idea would be to find the inverse mapping and use this inverse mapping to
  find how equations are transformed. E.g. if we let $z = x + iy$ and
  $f(z) = w = u + iv = u(x, y) + iv(x, y)$ then for part (a) the extended real
  axis in the domain is defined by $y = 0$. If there were an inverse mapping,
  then we could establish the following equation in the transformed plane:
  $\Im f^{-1} = 0$ and rearrange this equation to get an equation that
  describes the image of the extended real axis.

  However, the forward mapping is non-injective, so I don't think we can do that.

  Incidentally, I think the non-injectiveness and non-invertibility of the
  forward mapping can also be seen from this attempt to find the inverse, which leads
  to a quadratic expression with two solutions.
  \begin{align*}
    f(z)
    &= \frac{z^2 + 2z + 1}{z^2 - 2z + 1} \\
    &= w
  \end{align*}
  \begin{align*}
    z^2(1 - w) + 2z(1 + w) + (1 - w) = 0 \\
  \end{align*}
  \begin{align*}
    z &= \frac{-2(1 + w) \pm \sqrt{4(1 + w)^2 - 4(1 - w)^2}}{2(1 - w)} \\
      &= \frac{-(1 + w) \pm \sqrt{(1 + w)^2 - (1 - w)^2}}{1 - w} \\
      &= \frac{-(1 + w) \pm \sqrt{1 + 2w + w^2 - 1 + 2w - w^2}}{1 - w} \\
      &= \frac{-(1 + w) \pm 2\sqrt{w}}{1 - w} \\
  \end{align*}
\end{description}

\section{Elementary functions}

\subsection*{Exponential function}

Based on the premise that $(e^z)' = e^z$, we start with the Maclaurin series definition

\begin{align*}
  e^z = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + \ldots,
\end{align*}
which converges for all z.

Letting $z = x + iy$, we demand that $e^{x + iy} = e^xe^{iy}$, but what is $e^{iy}$?
\begin{align*}
  e^{iy}
  &= \sum_{n=0}^\infty \frac{i^ny^n}{n!} \\
  &= \sum_{k=0}^\infty (-1)^{k}\frac{y^{2k}}{(2k)!} +
     i\sum_{k=0}^\infty (-1)^{k}\frac{y^{2k + 1}}{(2k + 1)!}\\
\end{align*}
which are the Taylor series for $\cos$ and $\sin$. So
\begin{align*}
  e^{x + iy} = e^x(\cos y + i\sin y).
\end{align*}
The argument of the image point depends on the imaginary part of the input.

% \begin{itemize}
% \item inputs that have the same imaginary value will map to values with the
%   same argument: horizontal lines map to rays;
% \item horizontal lines offset by a multiple of $2\pi i$ will be mapped to the
%   same image ray;
% \item vertical lines map to concentric circles.
% \item The imaginary axis is wrapped round the unit circle.
% \item Any image point has infinitely many pre-image points; these differ by multiples of $2\pi i$.
% \end{itemize}

% The logarithm is the inverse of the exponential map $z \mapsto \exp(z)$, and
% therefore is multi-valued. So for example, the log of a point
% $\cos\theta + i\sin\theta$ on the unit circle could be one of infinitely many
% possibilities on the imaginary axis:
% $\ldots, i(\theta - 2\pi), i\theta, i(\theta + 2\pi),\ldots$.

Basically, the exponential map takes a vertical line $\Re z = x$ and wraps it
round a circle infinitely many times. The radius of the circle is $e^x$.

\subsection*{Branches of inverse functions}

A given point on the circle is hit by infinitely many points on the line:
$\ldots, x + i(y - 2\pi), x + iy, x + i(y + 2\pi), \ldots.$ These are the
logarithms of the point on the circle.

% The inverse of $\exp$ is $\log$. The $\log$ of a point $w$ is the set of points
% $z$ on the vertical line for which $e^z = w$.

The ``principal argument'' of a complex number $z$ is $\Arg z \in (-\pi,
\pi]$. It is continuous at points away from the negative real axis.

\includegraphics[width=200pt]{img/principal-arg.png}

The $n$-th roots of $z$ are
\begin{align*}
  \sqrt[n] {|z|}\left(\cos\left(\frac{\Arg z + 2k\pi}{n}\right) + i\sin\left(\frac{\Arg z + 2k\pi}{n}\right)\right),
\end{align*}
$k = 0, 1, \ldots, n-1$.
The ``principal root'' is given by $k = 0$:
\begin{align*}
  \sqrt[n] {|z|}\left(\cos\frac{\Arg z}{n} + i\sin\frac{\Arg z}{n}\right).
\end{align*}

The ``principal branch'' of the log function is given by
\begin{align*}
  \Log z = \ln |z| + i\Arg z.
\end{align*}

Because they involve $\Arg$, both $\Log$ and the principal root function are
continuous at points away from the negative real axis.

Here are some diagrams of the principal branch of the square root
function. Notice that it is discontinuous at points of the negative real
axis. I.e. it is a branch in the domain $\C \backslash [0, -\infty)$

\includegraphics[width=300pt]{img/principal-sqrt-2.png}
\includegraphics[width=200pt]{img/principal-sqrt.png}

Suppose we want a branch in a domain that includes the negative real axis. Then we can use
\begin{align*}
  g(z) =
  \begin{cases}
    \sqrt z,  &\Im z \geq 0\\
    -\sqrt z, &\Im z < 0
  \end{cases}
\end{align*}
and we could also use $-g$; there are two branches. However, these are
discontinuous at points on the positive real axis, so they are branches in the
domain $\C \backslash [0, +\infty)$.

Consider the infinitely wide strip $\{z: -\pi < \Im z < \pi\}$. The image of
this under the exponential map is the entire complex plane with 0 removed. For
example, consider some complex number $w$ with $|w| = r > 0$. It is hit by a
point on the vertical line $x = \ln r$. So this domain-restricted version of
the exponential map has an inverse, and that inverse is $\Log$.

For a multivalent function, it's not possible to find an inverse that sends
every image point $f(z)$ back to the correct place $z$ on the left-hand side,
for all $z$. That's because many $z$ hit the same $f(z)$. But it is possible to
find a ``right inverse'': a function that sends image points back to one of the
possible preimage points. We want this to be continuous, so we often have to
restrict the domain on the right-hand side to avoid points of discontinuity,
e.g. remove $(-\infty, 0]$ in the case of the $\Log$ and principal root inverse
functions.

\subsection*{Hyperbolic functions}

Just as their real counterparts,
\begin{align*}
  \cosh z &= (e^z + e^{-z})/2\\
  \sinh z &= (e^z - e^{-z})/2
\end{align*}
with
\begin{align*}
  \tanh &= \sinh/\cosh\\
  \coth &= 1/\tanh\\
  \sech &= 1/\cosh\\
  \cosech &= 1/\sinh
\end{align*}

\subsection*{Trigonometric functions}

From $e^{iy} = \cos y + i \sin y$ we have $e^{-iy} = \cos y - i \sin y$ and therefore
\begin{align*}
  \cos y &= (e^{iy} + e^{-iy})/2\\
  \sin y &= (e^{iy} - e^{-iy})/2i,
\end{align*}
for real $y$. $\cos$ and $\sin$ are defined on $\C$ by substituting a complex
variable $z$ in place of $y$:
\begin{align*}
  \cos z &= (e^{iz} + e^{-iz})/2\\
  \sin z &= (e^{iz} - e^{-iz})/2i,
\end{align*}
with $\tan, \cot, \sec$ and $\cosec$ also defined as usual.




\begin{description}
  \exercise{IV.5.2}{
    Describe the curves $|f| = $ constant and $\arg f =$ constant for the function
\begin{align*}
  f(z) = \exp(z^2).
\end{align*}
}
Let $z = x + iy$, so
\begin{align*}
  f(z)
  &= \exp((x^2 - y^2) + 2ixy) \\
  &= e^{x^2 - y^2}\(\cos 2xy + i\sin 2xy\)
\end{align*}
% Therefore the set of points $z = x + iy$ for which $|f(z)| = k \in \R$ are
% those satisfying $e^{x^2 - y^2} = k$.

Therefore $|f| = k$ for some constant $k \in \R$ implies that
$e^{x^2 - y^2} = k > 0$, i.e. $y = \pm \sqrt{x^2 - \log k}$. In other words,
the preimage of a circle of radius $k$ centered on the origin is the union of
the two curves $y = \pm \sqrt{x^2 - \log k}$.

$\arg f = \theta$ for some constant $0 \leq \theta < 2\pi$ implies that
$2xy = \theta$, i.e. $y = \frac{\theta}{2x}$. In other words, the preimage of a
ray at angle $\theta$ is the graph of $y = \frac{\theta}{2x}$.

\exercise{IV.9.2}{
  Find all values of $\log(\log i)$
} \\\\
$\log i$ is the following set of image points lying on the imaginary axis:
\begin{align*}
  \log i = \left\{i\(\frac{\pi}{2} + 2\pi k\):k \in \Z\right\}.
\end{align*}
Fix a particular $k$. The log of the corresponding image point is the following
set of secondary image points, lying on the vertical line through
$\frac{\pi}{2} + 2\pi k$:
\begin{align*}
  \log \(i\(\frac{\pi}{2} + 2\pi k\)\) = \left\{\log \(\frac{\pi}{2} + 2\pi k\) + i\(\frac{\pi}{2} + 2\pi l\):l \in \Z\right\},
\end{align*}
Therefore the set of all values of $\log(\log i)$ is the following rectangular grid of points
\begin{align*}
  \log \(\log i\) = \left\{\log \(\frac{\pi}{2} + 2\pi k\) + i\(\frac{\pi}{2} + 2\pi l\):k \in \Z, l \in \Z\right\}.
\end{align*}

\exercise{IV.13.3}{
  [Not in homework.] Let $G$ be the open set one obtains by removing from $\C$
  the interval $[-1, 1]$ on the real axis. Prove that there is a branch of the
  function $\sqrt{\frac{z+1}{z-1}}$ in G. (Suggestion: What is the image of $G$
  under the map $z \mapsto \frac{z+1}{z-1}$?)
} \\
The map $z \mapsto \frac{z+1}{z-1}$ maps points as follows:
\begin{align*}
  -1 &\mapsto 0  \\
  0 &\mapsto -1  \\
  1 &\mapsto \infty \\
  i &\mapsto -i \\
  -i &\mapsto i \\
\end{align*}
Thus
\begin{enumerate}
\item the image of $G$ is $\C$ with the negative real axis removed;
\item the image of the unit circle is the imaginary axis;
\item the image of the unit disc is the left half-plane and the image of the
  complement of the unit disc is the right half-plane.
\end{enumerate}


\exercise{IV.13.4}{ Let $G$ be as in Exercise IV.13.3. Prove that there is a
  branch of the function $\sqrt{z^2 - 1}$ in $G$.
} \\\\
Let $f(z) = \sqrt{z^2 - 1}$, using the principal square root function defined by
\begin{align*}
  \sqrt w = \sqrt{|w|}\left(\cos\frac{\Arg w}{2} + i\sin\frac{\Arg w}{2}\right).
\end{align*}
The principal square root function is discontinuous at points $w$ in
$(-\infty, 0]$. Therefore $f$ will be continuous for all $z \in \C$ except
where $z^2 - 1 \in (-\infty, 0]$, i.e.  $-1 \leq z \leq 1$. Therefore $f$ will
be continuous in $G = \C \backslash [-1, 1]$.


% A branch of the function $\sqrt{z^2 - 1}$ in $G$ is a function $\phi$ such that
% for every $z \in G$, $\sqrt{\phi(z)^2 + 1} = z$.

% Let $f$ be the map $z \mapsto z^2 - 1$.


$f$ maps points as follows:
\begin{align*}
  -1 &\mapsto 0  \\
  0 &\mapsto -1  \\
  1 &\mapsto 0 \\
  i &\mapsto -2 \\
  -i &\mapsto -2 \\
  \infty &\mapsto \infty \\
\end{align*}
Therefore the image of $G$ under $f$ is $\C$ with the interval $[-1, 0]$ removed.

% Suppose $f(z) = \sqrt[2^*]{z^2 - 1}$ where $\sqrt[2^*] w$ is the square root of
% $w$ with smallest positive argument. Then $f$ is not continuous in $G$ since if
% we travel round the excised interval, the root has changed sign by the time we
% return to the starting point. So $f$ is not a branch in $G$.

\exercise{IV.16.1}{
  Find all the values of $(1 + i)^i$.
} \\\\
$(1 + i)^i$ is the set of values
\begin{align*}
  \exp\left( i\log (1 + i) \right)
  &= \exp\left(i\left(\log\sqrt 2 + i\left(\frac{\pi}{4} + 2\pi k\right)\right)\right) \\
  &= \exp\(- \pi\left(2k + \frac{1}{4}\right) + i\frac{\log 2}{2}\) \\
  &= e^{- \pi\left(2k + \frac{1}{4}\right)}\left(\cos\frac{\log 2}{2} + i\sin\frac{\log 2}{2}\right)
\end{align*}
for $k \in \Z$.

\exercise{IV.16.3}{ Prove that if $f$ is a branch of $z^c$ in an open set not
  containing 0, then $f$ is holomorphic and $f'$ is a branch of $cz^{c-1}$.
} \\\\
A branch $f$ of $z^c$, defined on some open set excluding $0$, means that
$f(z) = e^{c\Log z}$ for some branch $\Log$ of the logarithm map.

% Similarly, a branch $g$ of $cz^{c-1}$, defined on the same open set excluding
% $0$, means that $g(z) = ce^{(c-1)\Log z}$.

The branch of the logarithm, the exponential function, and multiplication by a
complex number are all holomorphic transformations. Therefore $f$ is
holomorphic, because the composition of holomorphic functions with compatible
domains and ranges is holomorphic.

The derivative of $f$ is, by the chain rule,
\begin{align*}
f'(z) = c e^{c\Log z} \frac{d\Log z}{dz} = c e^{c\Log z} \frac{1}{z} = c \frac{f(z)}{z},
\end{align*}
where I have assumed without proof that $\frac{d\Log z}{dz} = \frac{1}{z}$. The
final expression above is a branch of $cz^{c-1}$ defined in the same region
that $f$ is defined.

\end{description}

\section{Power Series}

\begin{description}

  \exercise{V.6.2}{ Prove that the sequence $(g_n)_{n=0}^\infty$ converges
    locally uniformly in the open set $G$ if and only if it converges uniformly
    on each compact subset of $G$.
  }
  \\\\
  First, terminology: Sarason states that $(g_n)_{n=0}^\infty$ converges
  locally uniformly in $G$ if each point of $G$ has a neighborhood in which the
  sequence converges uniformly. I'm going to take that to mean ``if and only if''.

  For the forward direction, we need to show that if

  \emph{(A): each point of $G$ has a neighborhood in which $(g_n)$ converges uniformly}

  then

  \emph{(B): $(g_n)$ converges uniformly on each compact subset of $G$.}

  I don't have a proof, but a suggested approach for how to prove this is by contradiction:
  \begin{enumerate}
  \item Suppose (A) is true but that (B) is not, so that there exists some
    compact subset $S$ of $G$ on which $(g_n)$ does not converge uniformly.
  \item Show that there exists a point of $S$ which lacks any neighborhood
    within which convergence is uniform. $\qed$
  \end{enumerate}

  For the reverse direction, we need to show that if

  \emph{(B): $(g_n)$ converges uniformly on each compact subset of $G$.}

  then

  \emph{(A): each point of $G$ has a neighborhood in which $(g_n)$ converges uniformly}

  Again I don't have a proof, but a suggested approach for how to prove this is:
  \begin{enumerate}
  \item Consider a point $z$ of $G$.
  \item Show that there is a compact subset $S$ of $G$ that contains $z$.
  \item Show that $z$ has a neighborhood which is a subset of $S$. $\qed$
  \end{enumerate}

  \exercise{V.7.2}{
    Prove that the series
    $\sum_{n=0}^\infty \(\frac{z-1}{z+1}\)^n$ converges locally uniformly in the
    half-plane $\Re z > 0$, and find the sum.
  }\\
  (No attempt)

  \exercise{V.14.1(b)}{
    Find the radius of convergence of the following series:
    $$
    \sum_{n=0}^\infty \frac{(n!)^3}{(3n)!} z^{3n}
    $$
  }
  \\\\
  We use the ratio test:
\begin{align*}
  \limn \left|\frac{a_n}{a_{n+1}}\right|
  &= \limn \left|\frac{(n!)^3z^{3n}}{(3n)!}
                 \frac{(3n + 3)!}{((n+1)!)^3z^{3n + 3}}\right| \\
  &= |z^{-3}|\limn \left|\frac{(3n + 3)(3n + 2)(3n + 1)}{(n+1)^3}\right| \\
  &= |z^{-3}|\limn \left|\frac{27 + o(n^{-1})}{1 + o(n^{-1})}\right| \\
  &= 27|z^{-3}|
\end{align*}

So the series converges when $|z^3| > 27$, i.e. outside a disc of radius 3
centered at the origin. The radius of convergence is infinite.

\exercise{V.16.2}{
  What function is represented by the power series
  $\sum_{n=1}^\infty n^2z^n$?
}\\
(No attempt)

\exercise{V.18.1}{
  Use the scheme above to determine the power series with center 0 representing
  the function $f(z) = \frac{1}{1 + z + z^2}$ near 0. What is the radius of
  convergence of this series?
}\\\\

Assume $f(z)$ can be represented as a power series $\sum_{n=0}^\infty a_nz^n$. We
can write $f$ as the ratio
\begin{align*}
  f(z) = \frac{1}{1 + z + z^2} = \frac{\sum_{n=0}^\infty b_nz^n}{\sum_{n=0}^\infty c_nz^n} =: \frac{g(z)}{h(z)},
\end{align*}

where
\begin{align*}
  b_n &=
  \begin{cases}
    1, &n = 0 \\
    0, &\text{otherwise}
  \end{cases}\\
  c_n &=
  \begin{cases}
    1, & 0 \leq n \leq 2 \\
    0, &\text{otherwise}.
  \end{cases}
\end{align*}

Then
\begin{align*}
  g(z) = \sum_{n=0}^\infty b_nz^n
  &= f(z)h(z) \\
  &= \(\sum_{n=0}^\infty a_nz^n\)\(\sum_{n=0}^\infty c_nz^n\) \\
  &= \sum_{n=0}^\infty z^n \sum_{k=0}^n a_kc_{n-k} \\
\end{align*}
\end{description}

\section{Complex Integration}

\begin{description}
  \exercise{VI.7.2}{
    Derive the formula
    \begin{align*}
      \frac{1}{2\pi}\int_0^{2\pi} \cos^{2n} t ~dt = \frac{1 \cdot 3 \cdot 5 \cdots (2n - 1)}
                                                       {2 \cdot 4 \cdot 6 \cdots (2n)}
    \end{align*}
    by integrating the function $\frac{1}{z}\(z + \frac{1}{z}\)^{2n}$ around
    the unit circle, parameterized by the curve
    $\gamma(t) = e^{it} (0 \leq t \leq 2\pi)$.
  }
  \\
  Here are two slightly different attempts:

  \begin{mdframed}
    We can write the integral as
    \begin{align*}
    \frac{1}{2\pi}\int_0^{2\pi}\cos^{2n}t ~dt
    &= \frac{1}{2^{2n+1}\pi } \int_0^{2\pi} (e^{it} + e^{-it})^{2n} ~dt \\
    &= \frac{1}{2^{2n+1}\pi } \int_\gamma (z + z^{-1})^{2n} ~dz
  \end{align*}

  Now consider the related integral
  \begin{align*}
    \int_\gamma z^{-1}\(z + z^{-1}\)^{2n} ~dz
    &= \int_\gamma z^{-1}\sum_{k=0}^{2n} {2n \choose k} z^{2n - k}z^{-k} ~dz \\
    &= \sum_{k=0}^{2n} {2n \choose k} \int_\gamma z^{2(n - k) - 1} ~dz.
  \end{align*}
  If $k \neq n$, then $z^{2(n - k) - 1}$ is the derivative of
  $\frac{z^{2(n - k)}}{2(n - k)}$, in which case
  $\int_\gamma z^{2(n - k) - 1} ~dz = 0$ since $\gamma$ is a closed
  curve. Therefore the only terms remaining in the summation are those for
  which $k = n$:
  \begin{align*}
    \int_\gamma z^{-1}\(z + z^{-1}\)^{2n} ~dz
    &= {2n \choose n} \int_\gamma z^{-1} ~dz \\
    &= {2n \choose n} \int_0^{2\pi} e^{-it}ie^{it} ~dz \\
    &= {2n \choose n} 2\pi i.
  \end{align*}
  Returning to the original problem, we now know the value of a similar
  integral:
  \begin{align*}
    \frac{1}{2^{2n+1}\pi } \int_\gamma z^{-1}(z + z^{-1})^{2n} ~dz
    &= \frac{1}{2^{2n+1}\pi} {2n \choose n} 2\pi i \\
    &= \frac{1}{2^{2n+1}\pi} ~\frac{(2n)!}{2(n!)} ~2\pi i \\
    &= \frac{(n+1) \cdot (n+2) \cdots 2n}{2^{2n+1}} i \\
  \end{align*}


\end{mdframed}
~\\
~\\
\begin{mdframed}
  Alternatively we can write the integral as
  \begin{align*}
    \frac{1}{2\pi}\int_0^{2\pi}\cos^{2n}t ~dt
    &= \frac{1}{2^{2n+1}\pi } \int_0^{2\pi} (e^{it} + e^{-it})^{2n} ~dt \\
    &= \frac{1}{2^{2n+1}\pi } \int_\gamma (z + z^{-1})^{2n} ~dz \\
    &= \frac{1}{2^{2n+1}\pi } \int_\gamma \sum_{k=0}^{2n} {2n \choose k} z^{2n - k}z^{-k} ~dz \\
    &= \frac{1}{2^{2n+1}\pi } \sum_{k=0}^{2n} {2n \choose k} \int_\gamma z^{2(n - k)} ~dz.
  \end{align*}
  Now if $2n \neq k$, then $z^{2(n - k)}$ is the derivative of
  $\frac{z^{2(n - k) + 1}}{2(n - k) + 1}$, in which case
  $\int_\gamma z^{2(n - k)} ~dz = 0$.

  We can view the integral on the right side as integrating the function
  $(z + z^{-1})^{2n}$ around the unit circle:

\end{mdframed}
\exercise{VI.8.1}{
  Let $z_1$ and $z_2$ be distinct points of $\C$. Evaluate
  $\int_{[z1,z2]}z^n dz$ and $\int_{[z1,z2]}\bar{z}^n dz$ for
  $n = 0, 1, 2, \ldots$.
}
\\
\begin{mdframed}
  Let $\gamma(t) = z_1 + t(z_2 - z_1)$ for $t \in [0, 1]$ represent the curve $[z_1,z_2]$. We have
\begin{align*}
  \int_{[z1,z2]}z^n dz
  &= \int_0^1 \gamma(t)^n\gamma'(t) dt \\
  &= (z_2 - z_1)\int_0^1 \(z_1 + t(z_2 - z_1)\)^n dt. \\
  &= (z_2 - z_1)\sum_{k=0}^n {n \choose k} z_1^{n-k}(z_2 - z_1)^k \int_0^1 t^k dt \\
  &= \sum_{k=0}^n \frac{{n \choose k}}{k+1} z_1^{n-k}(z_2 - z_1)^{k+1}. \\
\end{align*}
And for $\bar z$ we have
\begin{align*}
  \int_{[z1,z2]}\bar{z}^n dz
  &= \int_0^1 \(\bar{\gamma(t)}\)^n\gamma'(t) dt \\
  &= (z_2 - z_1)\int_0^1 \(\bar{z_1} + t(\bar{z_2} - \bar{z_1})\)^n dt \\
  &= (z_2 - z_1)\sum_{k=0}^n {n \choose k} \bar{z_1}^{n-k}(\bar{z_2} - \bar{z_1})^k \int_0^1 t^k dt \\
  &= \sum_{k=0}^n \frac{{n \choose k}}{k+1} \bar{z_1}^{n-k}(z_2 - z_1)^{k+1} \\
\end{align*}


\end{mdframed}

\exercise{VI.8.3}{
  Let the complex-valued function $f$ be defined and continuous in the disc
  $|z-z_0| < R$. For $0 < r < R$ let $C_r$ denote the circle $|z - z_0| = r$,
  with counterclockwise orientation.
}
\\
\exercise{VI.8.4}{
  Assume that $f$ is of class $C^1$. Prove that
  \begin{align*}
    \lim_{r \rightarrow 0} \frac{1}{r^2} \int_{C_r} f(z) dz = 2\pi i \dfdzbar(z_0).
  \end{align*}
}
\\
\begin{mdframed}
  Let $\gamma(\theta) = z_0 + re^{i\theta}$ for $\theta \in [0, 2\pi]$ represent the curve $C_r$. Then
  \begin{align*}
    \lim_{r \rightarrow 0} \frac{1}{r^2} \int_{C_r} f(z) dz
    &= \lim_{r \rightarrow 0} \frac{1}{r^2} \int_0^{2\pi} f(\gamma(\theta))\gamma'(\theta) d\theta \\
    &= \lim_{r \rightarrow 0} \frac{i}{r} \int_0^{2\pi} f(z_0 + re^{i\theta})e^{i\theta} d\theta \\
    % &= f(z_0)\lim_{r \rightarrow 0} \frac{i}{r} \int_0^{2\pi} e^{i\theta} d\theta ~~~~\text{(should justify taking limit inside integral)}\\
  \end{align*}
  (Not sure where to go from here.)
  \begin{align*}
    \dfdzbar = \frac{1}{2}\(\ddx + i\ddy\)
  \end{align*}
\end{mdframed}
\exercise{VI.12.2}{
  Evaluate the integrals $\int_0^\infty \cos t^2 d\theta$
  and$\int_0^\infty \sin t^2 d\theta$ (the Fresnel intervals) by integrating
  $e^{-z^2}$ in the counterclockwise direction around the boundary of the
  region $\{z: |z|<R, 0 \leq \Arg z \leq \frac{\pi}{4}\}$ and letting
  $R \rightarrow \infty$.
}
\begin{mdframed}
  We represent the specified curve as $\gamma(\theta) = Re^{i\theta}$ for
  $\theta \in [0, \frac{\pi}{4}]$, in which case the specified integral is
  \begin{align*}
    \int_0^{\pi/4} e^{-e^{2i\theta}} Rie^{i\theta} d\theta =
    Ri\int_0^{\pi/4} e^{i\theta - e^{2i\theta}} d\theta.
  \end{align*}
\end{mdframed}

\exercise{}{Give an example of two convergent series whose product diverges.}
\end{description}