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 $z^n=r^ne^{i\theta n} = -1 \!\,$

$\Rightarrow r=1 \!\,$ and $i n \theta = i (\pi + 2\pi k ) \!\,$

$\Rightarrow \theta = \frac{\pi + 2\pi k}{n}\!\,$ , 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}

$\begin{align} Res(f,z_0)&= \left. \frac{1}{(z^n+1)' } \right|_{z=z_0} \\ &= \frac{1}{nz_0^{n-1}} \\ &= \frac{1}{ne^{\frac{i\pi}{n}(n-1)}} \\ &= \frac{e^{\frac{i\pi}{n}}}{ne^{i \pi}} \\ &= \frac{-1}{n}e^{\frac{i\pi}{n}} \end{align} \!\,$

Bound Arc Portion (B) of Integral

$\begin{align} |B| &=\left| \int_{S(R)} \frac{dz}{z^n+1} \right| \\ &\leq \int_{S(R)}\frac{dz}{|z^n+1|} \\ &= \int_{S(R)} \frac{dz}{|R^n+1|} \\ &\leq \frac{1}{R^n} \int_{S(R)} dz \\ &= \frac{1}{R^n} \frac{2 \pi}{n} R \\ &= \frac{2 \pi}{nR^{n-1}} \end{align}$

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 $dz=e^{i\frac{2\pi}{n}}dr \!\,$

$\begin{align} C &= \int_{Re^{i\frac{2\pi}{n}}}^0 \frac{dz}{1+z^n}\\ &= -\int_0^{Re^{i\frac{2\pi}{n}}}\frac{dz}{1+z^n} \\ &= -\int_0^R \frac{e^{i\frac{2\pi}{n}}}{1+(re^{i\frac{2\pi}{n}})^n}\\ &= -\int_0^R \frac{e^{i\frac{2\pi}{n}}}{1+r^n}\\ &= -e^{i\frac{2\pi}{n}} \underbrace{\int_0^R \frac{dr}{1+r^n}}_{A} \\ \end{align}$

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

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

We then have,

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

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 $\int\int_{|z|<1} |f(x+iy)|^2dx dy \leq 1 \!\,$ 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 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 $D(z_0,r_0) \!\, \in D$

$\begin{align} f^2(z_0) &= \frac{1}{2\pi i} \int_{|z-z_0|=r_0} \frac{f(z)}{z-z_0}dz \\ &= \frac{1}{2\pi i} \int_0^{2\pi} \frac{f^2(z_0+r_0e^{i\theta})ir_0e^{i\theta}}{(z_0+r_0e^{i\theta})-z_0} d\theta\\ &= \frac{1}{2\pi} \int_0^{2\pi} f^2(z_0+r_0e^{i\theta})d\theta \end{align}$

Integrate with respect to r

$\begin{align} \int_0^{r_0} r f^2(z_0)dr &= \int_0^{r_0} \int_0^{2\pi} f^2(z_0+re^{i\theta})rdr d\theta \\ &= \frac{1}{2\pi}\int\int_{D(z_0,r)}f^2(x+iy)dxdy \end{align}$

Integrating the left hand side, we have

$\int_0^{r_0} r f^2(z_0)dr = \frac{r_0^2}{2}f^2(z_0) \!\,$

Hence,

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

Bound |f(z0)| by using hypothesis

$\begin{align} \left| \frac{r_0^2}{2} f^2(z_0) \right| &= \left| \frac{1}{2 \pi} \int\int_{D(z_0,r_0)} f^2(x+iy)dxdy \right| \\ &\leq \frac{1}{2 \pi} \int\int_{D(z_0,r_0)} |f^2(x+iy)|dxdy \\ &\leq \frac{1}{2 \pi} \int\int_{D} |f^2(x+iy)|dxdy \\ &\leq \frac{1}{2\pi} \\ \\ \mbox{Then } \\ \left| \frac{r_0^2}{2} f^2(z_0) \right| &\leq \frac{1}{2 \pi} \\ \\ \mbox{This implies}\\ \\ |f(z_0)| &\leq \frac{1}{r_0 \sqrt{\pi}} \end{align}$

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