UMD Analysis Qualifying Exam/Aug06 Complex

Problem 2

For real $s$ consider the integral $\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{ist}}{t-i}dt.$ (a) Compute the Cauchy Principal Value of the integral (when it exists) (b) For which values of $s$ is the integral convergent?

Solution 2

Consider the complex function $f(z)=\frac{e^{isz}}{z-i}$. This function has a pole at $z=i$. We can calculate $\mathrm{Res}(f(z),i)=\underset{z\to i}{\lim }(z-i)f(z)=e^{-s}$.

Consider the contour ${\mathrm{\Gamma }}_R$ composed of the upper half circle $C_R$ centered at the origin with radius $R$ traversed counter-clockwise and the other part being the interval $[-R,R]$ on the real axis.

That is,

$\underset{{\mathrm{\Gamma }}_R}{\int }f(z)dz=\underset{C_R}{\int }f(z)dz+\underset{-R}{\overset{R}{\int }}f(t)dt=2\pi i\mathrm{Res}(f,i).$

Let us estimate the integral of $f$ along the half circle $C_R$. We parametrize $C_R$ by the path $z=R{e}^{i\theta }$, $dz=Ri{e}^{i\theta }$ for $\theta \in [0,\pi ]$. This gives

$\begin{array}{cc}|\underset{C_R}{\int }f(z)dz|=& \left|\underset{0}{\overset{\pi }{\int }}\frac{e^{isR{e}^{i\theta }}}{R{e}^{i\theta }-i}Ri{e}^{i\theta }d\theta \right|\\ \le & \underset{0}{\overset{\pi }{\int }}\left|e^{-Rs\sin (\theta )}eiRs\cos (\theta )\frac{Ri{e}^{i\theta }}{R{e}^{i\theta }-i}\right|d\theta \\ \le & \underset{0}{\overset{\pi }{\int }}{e}^{-Rs\sin (\theta )}\frac{R}{R-1}d\theta .\end{array}$

Break up the interval $[0,\pi ]$ into $[0,\delta ]\cup [\delta ,\pi -\delta ]\cup [\pi -\delta ,\pi ]$ for some $0<\delta <\pi /2$. This gives $\underset{0}{\overset{\pi }{\int }}{e}^{-Rs\sin (\theta )}\frac{R}{R-1}d\theta =\underset{\delta }{\overset{\pi -\delta }{\int }}{e}^{-Rs\sin (\theta )}\frac{R}{R-1}d\theta +2\underset{0}{\overset{\delta }{\int }}{e}^{-Rs\sin (\theta )}\frac{R}{R-1}d\theta$.

Let us evaluate the first of the two integrals on the right-hand side.

$\begin{array}{cc}\underset{\delta }{\overset{\pi -\delta }{\int }}{e}^{-Rs\sin (\theta )}\frac{R}{R-1}d\theta \le & \underset{\delta }{\overset{\pi -\delta }{\int }}{e}^{-Rs\sin (\delta )}\frac{R}{R-1}d\theta \\ =& (\pi -2\delta )e^{-Rs\sin (\delta )}\frac{R}{R-1}\end{array}$ which tends to 0 as $R\to \mathrm{\infty }$. NOTE: This argument only works if we assume $s>0$. If we try this argument for $s<0$, we bound the integrand by $e^{-Rs}\frac{R}{R-1}$ instead of $e^{-Rs\sin (\theta )}\frac{R}{R-1}$, but this will diverge as we send $R\to \mathrm{\infty }$ (which implies that $\underset{-R}{\overset{R}{\int }}f(t)dt$ must also diverge as $R\to \mathrm{\infty }$. This answers part b).

As for the other integral, $\begin{array}{c}2\underset{0}{\overset{\delta }{\int }}{e}^{-Rs\sin (\theta )}\frac{R}{R-1}d\theta \le 2\underset{0}{\overset{\delta }{\int }}\frac{R}{R-1}d\theta =2\delta \frac{R}{R-1}\end{array}$ which tends to $2\delta$ as $R\to \mathrm{\infty }$.

Therefore, we've shown that $\underset{R\to \mathrm{\infty }}{\lim }|\underset{C_R}{\int }f(z)dz|\le 2\delta$. But $0<\delta <\pi /2$ was arbitrary, hence we can say that the integral vanishes.

Therefore, $2\pi i{e}^{-s}=\underset{R\to \mathrm{\infty }}{\lim }(\underset{C_R}{\int }f(z)dz+\underset{-R}{\overset{R}{\int }}f(t)dt)=0+\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}\frac{e^{ist}}{t-i}dt.$

Problem 4

Let $D=\{|z|<1\}$ have boundary $S=\{|z|=1\}$. For $\zeta \in D$ define $f(z)=\frac{\zeta -z^2}{1-\overline{\zeta }{z}^2}$. (a) Show that $f(z)\in S$ if and only if $z\in S$. (b) Show that $f$ has at least one fixed point $\omega \in S$.

Solution 4

4a

Consider $g(z)=\frac{\zeta -z}{1-\overline{\zeta }z}$ and $h(z)=z^2$. Then $f(z)=g\circ h(z)$. We know that $h(z)$ is a conformal map from $D$ to $D$ and moreover, $f(z)\in S$ if an only if $z\in S$. The same is true for $h(z)$, that is, $h(z)=z^2\in S$ if any only if $z\in S$. Therefore, $f(z)=g\circ h(z)$ if and only if $z\in S$.

4b

If $\omega$ is a fixed point of $f(z)$, then $f(\omega )=\frac{\zeta -{\omega }^2}{1-\overline{\zeta }{\omega }^2}=\omega$. Rearranging gives $\bar{\zeta }{\omega }^3-{\omega }^2-\omega +\zeta .$ By the fundamental theorem of algebra, we are guaranteed 3 solutions to this equation in the complex plane. All that we need to show is that at least on of these solutions lie on the circle n the circle $|\omega |=1$.

Problem 6

Let $ℱ$ be a family of entire functions. For $n=0,\pm 1,\pm 2,...$ define the domains $D_n=\{n-2<Re(z)<n+2\}$. If $ℱ$ is normal (i.e. convergence to $\mathrm{\infty }$ is allowed) on each $D_n$ show that $ℱ$ is normal on $ℂ$.