UMD Analysis Qualifying Exam/Aug06 Complex

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

$$\begin{align} \left|\int_{C_R}f(z)\,dz\right| =& \left| \int_0^\pi \frac{e^{isRe^{i\theta}}}{Re^{i\theta}-i} R i e^{i\theta}\,d\theta \right|\\ \leq & \int_0^\pi \left| e^{-R s \sin(\theta)} e{iRs\cos(\theta)} \frac{Rie^{i\theta}}{Re^{i\theta}-i}\right|\,d\theta\\ \leq & \int_0^\pi e^{-Rs\sin(\theta)} \frac{R}{R-1} \,d\theta. \end{align}$$

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.

$$\begin{align} \int_\delta^{\pi-\delta} e^{-Rs\sin(\theta)} \frac{R}{R-1} \,d\theta \leq & \int_\delta^{\pi-\delta} e^{-Rs\sin(\delta)} \frac{R}{R-1} \,d\theta\\ =& (\pi - 2\delta) e^{-R s\sin(\delta)} \frac{R}{R-1} \end{align}$$ 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, $$\begin{align} 2\int_0^\delta e^{-Rs\sin(\theta)} \frac{R}{R-1} \,d\theta \leq 2 \int_0^\delta \frac{R}{R-1} \,d\theta = 2\delta \frac{R}{R-1} \end{align}$$ 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, $$2\pi i e^{-s}=\lim_{R\to\infty} (\int_{C_R}f(z)\,dz + \int_{-R}^R f(t)\,dt )= 0+\int_{-\infty}^\infty \frac{e^{ist}}{t-i}\,dt.$$

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 $$|\omega|=1$$.