UMD Analysis Qualifying Exam/Aug08 Complex

Problem 2

Compute $\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^3+1}dx$

Solution 2

We will compute the general case:

$\underset{0}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^n+1}dx$

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 $z=r{e}^{i\theta }$ 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 $\mathrm{\infty }$ , 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

$\underset{\gamma }{\int }f(z)=\underset{A}{\underbrace{\underset{0}{\overset{R}{\int }}f(z)}}+\underset{B}{\underbrace{\underset{S(R)}{\int }f(z)}}+\underset{C}{\underbrace{\underset{R{e}^{i\frac{2\pi }{n}}}{\overset{0}{\int }}f(z)}}$

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

$\begin{array}{cc}Res(f,z_0)& ={\frac{1}{(z^n+1{)}^{\prime }}|}_{z=z_0}\\ & =\frac{1}{n{z}_0^{n-1}}\\ & =\frac{1}{n{e}^{\frac{i\pi }{n}(n-1)}}\\ & =\frac{e^{\frac{i\pi }{n}}}{n{e}^{i\pi }}\\ & =\frac{-1}{n}{e}^{\frac{i\pi }{n}}\end{array}$

Bound Arc Portion (B) of Integral

$\begin{array}{cc}|B|& =\left|\underset{S(R)}{\int }\frac{dz}{z^n+1}\right|\\ & \le \underset{S(R)}{\int }\frac{dz}{|z^n+1|}\\ & =\underset{S(R)}{\int }\frac{dz}{|R^n+1|}\\ & \le \frac{1}{R^n}\underset{S(R)}{\int }dz\\ & =\frac{1}{R^n}\frac{2\pi }{n}R\\ & =\frac{2\pi }{n{R}^{n-1}}\end{array}$

Hence as $R\to \mathrm{\infty }$, $|B|\to 0$

Parametrize (C) in terms of (A)

Let $z=r{e}^{i\frac{2\pi }{n}}$ where $r$ is real number. Then $dz=e^{i\frac{2\pi }{n}}dr$

$\begin{array}{cc}C& =\underset{R{e}^{i\frac{2\pi }{n}}}{\overset{0}{\int }}\frac{dz}{1+z^n}\\ & =-\underset{0}{\overset{R{e}^{i\frac{2\pi }{n}}}{\int }}\frac{dz}{1+z^n}\\ & =-\underset{0}{\overset{R}{\int }}\frac{e^{i\frac{2\pi }{n}}}{1+(r{e}^{i\frac{2\pi }{n}}{)}^n}\\ & =-\underset{0}{\overset{R}{\int }}\frac{e^{i\frac{2\pi }{n}}}{1+r^n}\\ & =-e^{i\frac{2\pi }{n}}\underset{A}{\underbrace{\underset{0}{\overset{R}{\int }}\frac{dr}{1+r^n}}}\\ \end{array}$

Apply Cauchy Integral Formula

From Cauchy Integral Formula, we have,

$A+B+C=2\pi i\frac{-e^{i\frac{\pi }{n}}}{n}$

As $R\to \mathrm{\infty }$, $B\to 0$. Also $C$ can be written in terms of $A$. Hence

$\begin{array}{cc}A+B+C& =A+C\\ & =(1-e^{i\frac{2\pi }{n}})A\end{array}$

We then have,

$\begin{array}{cc}A& =\frac{2\pi i}{n}\frac{e^{i\frac{\pi }{n}}}{e^{i\frac{2\pi }{n}}-1}\\ & =\frac{\pi }{n}\frac{2i{e}^{i\frac{\pi }{n}}}{e^{i\frac{\pi }{n}}(e^{i\frac{\pi }{n}}-e^{-i\frac{\pi }{n}})}\\ & =\frac{\pi }{n}\frac{1}{\sin (\frac{\pi }{n})}\end{array}$

Problem 4

Suppose $S=\{z:-\frac{\pi }{2}<\mathrm{\Im }(z)<\frac{\pi }{2}\}$ 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 $h:D\to D$ be the automorphism defined as

$h(z)=\frac{a-z}{1-\bar{a}z}$

Consider now $F(z)=h\circ f\circ h^{-1}(z)$. Then, F has two fixed points, namely

$F(0)=h\circ f\circ h^{-1}(0)=h\circ f(a)=h(a)=0$

$F\left(\frac{a-b}{1-\bar{a}b}\right)=h\circ f\circ h^{-1}\left(\frac{a-b}{1-\bar{a}b}\right)=h\circ f(b)=h(b)=\frac{a-b}{1-\bar{a}b}$.

Since $F(0)=0$,

$\frac{a-b}{1-\bar{a}b}\ne 0$ (since $a$ is different to $b$), and

$\left|F\left(\frac{a-b}{1-\bar{a}b}\right)\right|=\left|\frac{a-b}{1-\bar{a}b}\right|$,

by Schwarz Lemma,

$F(z)=\alpha z$.

But, replacing $\frac{a-b}{1-\bar{a}b}$ 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 $ℱ$ be the family of functions $f$ analytic on $\{|z|<1\}$ so that $\int \underset{|z|<1}{\int }|f(x+iy){|}^{2}dxdy\le 1$ Prove that $ℱ$ 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 $f\in ℱ$ 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, $|f(z_0)|=\underset{z\in K}{\max }|f(z|$.

Note that $z_0$ is independent of the choice of $f\in ℱ$.

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 $D(z_0,r_0)\in D$

$\begin{array}{cc}{f}^2(z_0)& =\frac{1}{2\pi i}\underset{|z-z_0|=r_0}{\int }\frac{f^2(z)}{z-z_0}dz\\ & =\frac{1}{2\pi i}\underset{0}{\overset{2\pi }{\int }}\frac{f^2(z_0+r_0{e}^{i\theta })i{r}_0{e}^{i\theta }}{(z_0+r_0{e}^{i\theta })-z_0}d\theta \\ & =\frac{1}{2\pi }\underset{0}{\overset{2\pi }{\int }}{f}^2(z_0+r_0{e}^{i\theta })d\theta \end{array}$

Integrate with respect to r

$\begin{array}{cc}\underset{0}{\overset{r_0}{\int }}r{f}^2(z_0)dr& =\underset{0}{\overset{r_0}{\int }}\underset{0}{\overset{2\pi }{\int }}{f}^2(z_0+r{e}^{i\theta })rdrd\theta \\ & =\frac{1}{2\pi }\int \underset{D(z_0,r)}{\int }{f}^2(x+iy)dxdy\end{array}$

Integrating the left hand side, we have

$\underset{0}{\overset{r_0}{\int }}r{f}^2(z_0)dr=\frac{r_0^2}{2}{f}^2(z_0)$

Hence,

$\frac{r_0^2}{2}{f}^2(z_0)=\frac{1}{2\pi }\int \underset{D(z_0,r)}{\int }{f}^2(x+iy)dxdy$

Bound |f(z0)| by using hypothesis

$\begin{array}{cc}|\frac{r_0^2}{2}{f}^2(z_0)|& =|\frac{1}{2\pi }\int \underset{D(z_0,r_0)}{\int }{f}^2(x+iy)dxdy|\\ & \le \frac{1}{2\pi }\int \underset{D(z_0,r_0)}{\int }|f^2(x+iy)|dxdy\\ & \le \frac{1}{2\pi }\int \underset{D}{\int }|f^2(x+iy)|dxdy\\ & \le \frac{1}{2\pi }\\ \\ \text{Then }\\ |\frac{r_0^2}{2}{f}^2(z_0)|& \le \frac{1}{2\pi }\\ \\ \text{This implies}\\ \\ |f(z_0)|& \le \frac{1}{r_0\sqrt{\pi }}\end{array}$

Apply Montel's Theorem

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