UMD Analysis Qualifying Exam/Aug06 Complex

Problem 2

For real $s$ consider the integral

$\int _{-\infty }^{\infty }{\frac {e^{ist}}{t-i}}\,dt.$

(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 $\operatorname {Res} (f(z),i)=\lim _{z\to i}(z-i)f(z)=e^{-s}$ .

Consider the contour $\Gamma _{R}$ 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,

$\int _{\Gamma _{R}}f(z)\,dz=\int _{C_{R}}f(z)\,dz+\int _{-R}^{R}f(t)\,dt=2\pi i\operatorname {Res} (f,i).$

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

${\begin{aligned}\left|\int _{C_{R}}f(z)\,dz\right|=&\left|\int _{0}^{\pi }{\frac {e^{isRe^{i\theta }}}{Re^{i\theta }-i}}Rie^{i\theta }\,d\theta \right|\\\leq &\int _{0}^{\pi }\left|e^{-Rs\sin(\theta )}e{iRs\cos(\theta )}{\frac {Rie^{i\theta }}{Re^{i\theta }-i}}\right|\,d\theta \\\leq &\int _{0}^{\pi }e^{-Rs\sin(\theta )}{\frac {R}{R-1}}\,d\theta .\end{aligned}}$

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 $\int _{0}^{\pi }e^{-Rs\sin(\theta )}{\frac {R}{R-1}}\,d\theta =\int _{\delta }^{\pi -\delta }e^{-Rs\sin(\theta )}{\frac {R}{R-1}}\,d\theta +2\int _{0}^{\delta }e^{-Rs\sin(\theta )}{\frac {R}{R-1}}\,d\theta$ .

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

${\begin{aligned}\int _{\delta }^{\pi -\delta }e^{-Rs\sin(\theta )}{\frac {R}{R-1}}\,d\theta \leq &\int _{\delta }^{\pi -\delta }e^{-Rs\sin(\delta )}{\frac {R}{R-1}}\,d\theta \\=&(\pi -2\delta )e^{-Rs\sin(\delta )}{\frac {R}{R-1}}\end{aligned}}$ which tends to 0 as $R\to \infty$ . 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 $R\to \infty$ (which implies that $\int _{-R}^{R}f(t)\,dt$ must also diverge as $R\to \infty$ . This answers part b).

As for the other integral, ${\begin{aligned}2\int _{0}^{\delta }e^{-Rs\sin(\theta )}{\frac {R}{R-1}}\,d\theta \leq 2\int _{0}^{\delta }{\frac {R}{R-1}}\,d\theta =2\delta {\frac {R}{R-1}}\end{aligned}}$ which tends to $2\delta$ as $R\to \infty$ .

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

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

Problem 4

Let $D=\{|z|<1\}$ have boundary $S=\{|z|=1\}$ . For $\zeta \in D$ define

$f(z)={\frac {\zeta -z^{2}}{1-{\bar {\zeta }}z^{2}}}$ .

(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 $g(z)={\frac {\zeta -z}{1-{\bar {\zeta }}z}}$ 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 $f(\omega )={\frac {\zeta -\omega ^{2}}{1-{\bar {\zeta }}\omega ^{2}}}=\omega$ . Rearranging gives ${\overline {\zeta }}\omega ^{3}-\omega ^{2}-\omega +\zeta .$ 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 $|\omega |=1$ .