# UMD Analysis Qualifying Exam/Aug08 Complex
## Problem 2

* Compute
* $\int _{0}^{\infty }{\frac {1}{x^{3}+1}}dx\!\,$

## Solution 2

We will compute the general case:
$\int _{0}^{\infty }{\frac {1}{x^{n}+1}}dx\!\,$

### Find Poles of f(z)

The poles of $f(z)={\frac {1}{z^{n}+1}}\!\,$ are just the zeros of $z^{n}+1\!\,$, so we can compute them in the following manner:
If $z=re^{i\theta }\!\,$ is a solution of $z^{n}+1=0\!\,$,
then $z^{n}=r^{n}e^{i\theta n}=-1\!\,$
$\Rightarrow r=1\!\,$ and $in\theta =i(\pi +2\pi k)\!\,$
$\Rightarrow \theta ={\frac {\pi +2\pi k}{n}}\!\,$ , k=0,1,2,...,n-1.
Thus, the poles of $f(z)\!\,$ are of the form $z=e^{\frac {\pi +2\pi k}{n}}\!\,$ with $k=0,1,...,n-1\!\,$

### Choose Path of Contour Integral

In order to get obtain the integral of $f(x)\!\,$ from 0 to $\infty \!\,$ , let us consider the path $\gamma \!\,$ consisting in a line $A\!\,$ going from 0 to $R\!\,$, then the arc $B\!\,$ of radius $R\!\,$ from the angle 0 to ${\frac {2\pi }{n}}\!\,$ and then the line $C\!\,$ joining the end point of $B\!\,$ and the initial point of $A\!\,$,
where $R\!\,$ is a fixed positive number such that
the pole $z_{0}=e^{i{\frac {\pi }{n}}}\!\,$ is inside the curve $\gamma \!\,$. Then , we need to estimate the integral
$\int _{\gamma }f(z)=\underbrace {\int _{0}^{R}f(z)} _{A}+\underbrace {\int _{S(R)}f(z)} _{B}+\underbrace {\int _{Re^{i{\frac {2\pi }{n}}}}^{0}f(z)} _{C}\!\,$

### Compute Residues of f at z0= exp{i\pi /n}

${\begin{aligned}Res(f,z_{0})&=\left.{\frac {1}{(z^{n}+1)'}}\right|_{z=z_{0}}\\&={\frac {1}{nz_{0}^{n-1}}}\\&={\frac {1}{ne^{{\frac {i\pi }{n}}(n-1)}}}\\&={\frac {e^{\frac {i\pi }{n}}}{ne^{i\pi }}}\\&={\frac {-1}{n}}e^{\frac {i\pi }{n}}\end{aligned}}\!\,$

### Bound Arc Portion (B) of Integral

${\begin{aligned}|B|&=\left|\int _{S(R)}{\frac {dz}{z^{n}+1}}\right|\\&\leq \int _{S(R)}{\frac {dz}{|z^{n}+1|}}\\&=\int _{S(R)}{\frac {dz}{|R^{n}+1|}}\\&\leq {\frac {1}{R^{n}}}\int _{S(R)}dz\\&={\frac {1}{R^{n}}}{\frac {2\pi }{n}}R\\&={\frac {2\pi }{nR^{n-1}}}\end{aligned}}$
Hence as $R\rightarrow \infty \!\,$, $|B|\rightarrow 0\!\,$

### Parametrize (C) in terms of (A)

Let $z=re^{i{\frac {2\pi }{n}}}\!\,$ where $r\!\,$ is real number. Then $dz=e^{i{\frac {2\pi }{n}}}dr\!\,$
${\begin{aligned}C&=\int _{Re^{i{\frac {2\pi }{n}}}}^{0}{\frac {dz}{1+z^{n}}}\\&=-\int _{0}^{Re^{i{\frac {2\pi }{n}}}}{\frac {dz}{1+z^{n}}}\\&=-\int _{0}^{R}{\frac {e^{i{\frac {2\pi }{n}}}}{1+(re^{i{\frac {2\pi }{n}}})^{n}}}\\&=-\int _{0}^{R}{\frac {e^{i{\frac {2\pi }{n}}}}{1+r^{n}}}\\&=-e^{i{\frac {2\pi }{n}}}\underbrace {\int _{0}^{R}{\frac {dr}{1+r^{n}}}} _{A}\\\end{aligned}}$

### Apply Cauchy Integral Formula

From Cauchy Integral Formula, we have,
    $A+B+C=2\pi i{\frac {-e^{i{\frac {\pi }{n}}}}{n}}\!\,$
As $R\rightarrow \infty \!\,$, $B\rightarrow 0\!\,$. Also $C\!\,$ can be written in terms of $A\!\,$. Hence
${\begin{aligned}A+B+C&=A+C\\&=(1-e^{i{\frac {2\pi }{n}}})A\end{aligned}}$
We then have,
${\begin{aligned}A&={\frac {2\pi i}{n}}{\frac {e^{i{\frac {\pi }{n}}}}{e^{i{\frac {2\pi }{n}}}-1}}\\&={\frac {\pi }{n}}{\frac {2ie^{i{\frac {\pi }{n}}}}{e^{i{\frac {\pi }{n}}}(e^{i{\frac {\pi }{n}}}-e^{-i{\frac {\pi }{n}}})}}\\&={\frac {\pi }{n}}{\frac {1}{\sin({\frac {\pi }{n}})}}\end{aligned}}$

## Problem 4

* Suppose $S=\{z:-{\frac {\pi }{2}}<\Im (z)<{\frac {\pi }{2}}\}\!\,$ and there is an entire function $g\!\,$ with $g(S)\subset S\!\,$. If $g(-1)=0\!\,$ and $g(0)=1\!\,$, prove that $g(z)=z+1\!\,$

## Solution 4
### Lemma: Two fixed points imply identity

Lemma. Let $f\!\,$ be analytic on the unit $D\!\,$, and assume that $|f(z)|<1\!\,$ on the disc. Prove that if there exist two distinct points $a\!\,$ and $b\!\,$ in the disc which are fixed points, that is, $f(a)=a\!\,$ and $f(b)=b\!\,$, then $f(z)=z\!\,$.
Proof Let $h:D\rightarrow D\!\,$ be the automorphism defined as
$h(z)={\frac {a-z}{1-{\overline {a}}z}}\!\,$
Consider now $F(z)=h\circ f\circ h^{-1}(z)\!\,$. Then, F has two fixed points, namely
$F(0)=h\circ f\circ h^{-1}(0)=h\circ f(a)=h(a)=0\!\,$
$F\left({\frac {a-b}{1-{\overline {a}}b}}\right)=h\circ f\circ h^{-1}\left({\frac {a-b}{1-{\overline {a}}b}}\right)=h\circ f(b)=h(b)={\frac {a-b}{1-{\overline {a}}b}}\!\,$.
Since $F(0)=0\!\,$,
${\frac {a-b}{1-{\overline {a}}b}}\neq 0\!\,$ (since $a\!\,$ is different to $b\!\,$), and
$\left|F\left({\frac {a-b}{1-{\overline {a}}b}}\right)\right|=\left|{\frac {a-b}{1-{\overline {a}}b}}\right|\!\,$,
by Schwarz Lemma,
$F(z)=\alpha z\!\,$.
But, replacing ${\frac {a-b}{1-{\overline {a}}b}}\!\,$ into the last formula, we get $\alpha =1\!\,$.
Therefore,
$h\circ f\circ h^{-1}(z)=z\!\,$,
which implies
$f(z)=z\!\,$

### Shift Points to Create Fixed Points

Let $f(z)=g(z)-1\!\,$. Then $f(0)=0\!\,$ and $f(-1)=-1\!\,$.
Notice that $S\!\,$ is an infinite horizontal strip centered around the real axis with height $\pi \!\,$. Since $f(z)\!\,$ is a unit horizontal shift left, $f(S)\subset S\!\,$.

### Use Riemann Mapping Theorem

From the Riemann mapping theorem, there exists a biholomorphic (bijective and holomorphic) mapping $h\!\,$, from the open unit disk $D\!\,$ to $S\!\,$.

### Define Composition Function

Let $F=h^{-1}\circ f\circ h\!\,$. Then $F\!\,$ maps $D\!\,$ to $D\!\,$.
From the lemma, since $F(z)\!\,$ has two fixed points, $F(z)=z\!\,$ which implies $f(z)=z\!\,$ which implies $g=z+1\!\,$.

## Problem 6

* Let ${\mathcal {F}}\!\,$ be the family of functions $f\!\,$ analytic on $\{|z|<1\}\!\,$ so that
* $\int \int _{|z|<1}|f(x+iy)|^{2}dxdy\leq 1\!\,$
* Prove that ${\mathcal {F}}\!\,$ is a normal family on $\{|z|<1\}\!\,$

## Solution 6
### Choose any compact set K in D

Choose any compact set $K\!\,$ in the open unit disk $D\!\,$. Since $K\!\,$ is compact, it is also closed and bounded.
We want to show that for all $f\in {\mathcal {F}}\!\,$ and all $z\in K\!\,$, $|f(z)|\!\,$ is bounded i.e.
    $|f(z)|<B_{K}\!\,$
where $B_{K}\!\,$ is some constant dependent on the choice of $K\!\,$.

### Apply Maximum Modulus Principle to find |f(z0)|

Choose $z_{0}\!\,$ that is the shortest distance from the boundary of the unit disk $D\!\,$. From the maximum modulus principle, $|f(z_{0})|=\max _{z\in K}|f(z|\!\,$.
Note that $z_{0}\!\,$ is independent of the choice of $f\in {\mathcal {F}}\!\,$.

### Apply Cauchy's Integral Formula to f^2(z0)

We will apply Cauchy's Integral formula to $f^{2}(z_{0})\!\,$ (instead of $f(z_{0})\!\,$) to take advantage of the hypothesis.
Choose sufficiently small $r_{0}>0\!\,$ so that $D(z_{0},r_{0})\!\,\in D$
${\begin{aligned}f^{2}(z_{0})&={\frac {1}{2\pi i}}\int _{|z-z_{0}|=r_{0}}{\frac {f^{2}(z)}{z-z_{0}}}dz\\&={\frac {1}{2\pi i}}\int _{0}^{2\pi }{\frac {f^{2}(z_{0}+r_{0}e^{i\theta })ir_{0}e^{i\theta }}{(z_{0}+r_{0}e^{i\theta })-z_{0}}}d\theta \\&={\frac {1}{2\pi }}\int _{0}^{2\pi }f^{2}(z_{0}+r_{0}e^{i\theta })d\theta \end{aligned}}$

### Integrate with respect to r

${\begin{aligned}\int _{0}^{r_{0}}rf^{2}(z_{0})dr&=\int _{0}^{r_{0}}\int _{0}^{2\pi }f^{2}(z_{0}+re^{i\theta })rdrd\theta \\&={\frac {1}{2\pi }}\int \int _{D(z_{0},r)}f^{2}(x+iy)dxdy\end{aligned}}$
Integrating the left hand side, we have
$\int _{0}^{r_{0}}rf^{2}(z_{0})dr={\frac {r_{0}^{2}}{2}}f^{2}(z_{0})\!\,$
Hence,
${\frac {r_{0}^{2}}{2}}f^{2}(z_{0})={\frac {1}{2\pi }}\int \int _{D(z_{0},r)}f^{2}(x+iy)dxdy\!\,$

### Bound |f(z0)| by using hypothesis

${\begin{aligned}\left|{\frac {r_{0}^{2}}{2}}f^{2}(z_{0})\right|&=\left|{\frac {1}{2\pi }}\int \int _{D(z_{0},r_{0})}f^{2}(x+iy)dxdy\right|\\&\leq {\frac {1}{2\pi }}\int \int _{D(z_{0},r_{0})}|f^{2}(x+iy)|dxdy\\&\leq {\frac {1}{2\pi }}\int \int _{D}|f^{2}(x+iy)|dxdy\\&\leq {\frac {1}{2\pi }}\\\\{\mbox{Then }}\\\left|{\frac {r_{0}^{2}}{2}}f^{2}(z_{0})\right|&\leq {\frac {1}{2\pi }}\\\\{\mbox{This implies}}\\\\|f(z_{0})|&\leq {\frac {1}{r_{0}{\sqrt {\pi }}}}\end{aligned}}$

### Apply Montel's Theorem

Then, since any $f\in {\mathcal {F}}\!\,$ is uniformly bounded in every compact set, by Montel's Theorem, it follows that ${\mathcal {F}}\!\,$ is normal