# UMD Analysis Qualifying Exam/Aug06 Complex

## Problem 2

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

### Solution 2

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

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

That is,

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

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

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

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

{\displaystyle {\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 ${\displaystyle R\to \infty }$ . NOTE: This argument only works if we assume ${\displaystyle s>0}$ . If we try this argument for ${\displaystyle s<0}$ , we bound the integrand by ${\displaystyle e^{-Rs}{\frac {R}{R-1}}}$  instead of ${\displaystyle e^{-Rs\sin(\theta )}{\frac {R}{R-1}}}$ , but this will diverge as we send ${\displaystyle R\to \infty }$  (which implies that ${\displaystyle \int _{-R}^{R}f(t)\,dt}$  must also diverge as ${\displaystyle R\to \infty }$ . This answers part b).

As for the other integral, {\displaystyle {\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 ${\displaystyle 2\delta }$  as ${\displaystyle R\to \infty }$ .

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

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

## Problem 4

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

### Solution 4

#### 4a

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

#### 4b

If ${\displaystyle \omega }$  is a fixed point of ${\displaystyle f(z)}$ , then ${\displaystyle f(\omega )={\frac {\zeta -\omega ^{2}}{1-{\bar {\zeta }}\omega ^{2}}}=\omega }$ . Rearranging gives ${\displaystyle {\overline {\zeta }}\omega ^{3}-\omega ^{2}-\omega +\zeta .}$  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 ${\displaystyle |\omega |=1}$ .

## Problem 6

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