91Ó°ÊÓ

Find the location of the branch points and discuss possible branch cuts for the following functions: (a) \(\frac{1}{(z-1)^{1 / 2}}\) (b) \((z+1-2 i)^{1 / 4}\) (c) \(2 \log z^{2}\) (d) \(z^{\sqrt{2}}\)

Determine whether the following functions are analytic. Discuss whether they have any singular points or if they are entire. (a) \(\tan z\) (b) \(e^{\sin z}\) (c) \(e^{1 /(z-1)}\) (d) \(e^{\bar{z}}\) (e) \(\frac{z}{z^{4}+1}\) (f) \(\cos x \cosh y-i \sin x \sinh y\)

We wish to evaluate the integral \(I=\int_{0}^{\infty} e^{i x^{2}} d x\). Consider the contour \(I_{R}=\oint_{\left.C_{i R}\right)} e^{i z^{2}} d z\), where \(C_{(R)}\) is the closed circular sector in the upper half plane with boundary points \(0, R\), and \(R e^{i \pi / 4}\). Show that \(I_{R}=0\) and that \(\lim _{R \rightarrow \infty} \int_{C_{1(k)}} e^{i z^{2}} d z=0\), where \(C_{1(R)}\) is the line integral along the circular sector from \(R\) to \(R e^{i \pi / 4}\). (Hint: use \(\sin x \geq \frac{2 x}{\pi}\) on \(0 \leq x \leq \frac{\pi}{2} .\) ) Then, breaking up the contour \(C_{(R)}\) into three component parts, deduce $$ \lim _{R \rightarrow \infty}\left(\int_{0}^{R} e^{i x^{2}} d x-e^{i \pi / 4} \int_{0}^{R} e^{-r^{2}} d r\right)=0 $$ and from the well-known result of real integration, \(\int_{0}^{\infty} e^{-x^{2}} d x=\sqrt{\pi} / 2\), deduce that \(I=e^{i \pi / 4} \sqrt{\pi} / 2\)

Derive the following formulae: (a) \(\operatorname{coth}^{-1} z=\frac{1}{2} \log \frac{z+1}{z-1}\) (b) \(\operatorname{sech}^{-1} z=\log \left(\frac{1+\left(1-z^{2}\right)^{1 / 2}}{z}\right)\)

Let \(f(z)\) be analytic in some domain. Show that \(f(z)\) is necessarily a constant if either the function \(\overline{f(z)}\) is analytic or \(f(z)\) assumes only pure imaginary values in the domain.

Consider the integral \(\int_{0}^{b}\left(1 / z^{1 / 2}\right) d z, b>0\). Let \(z^{1 / 2}\) have a branch cut along the positive real axis. Show that the value of the integral obtained by integrating along the top half of the cut is exactly minus that obtained by integrating along the bottom half of the cut. What is the difference between taking the principal versus the second branch of \(z^{1 / 2} ?\)

Let \(f(z)\) be an entire function, with \(|f(z)| \leq C|z|\) for all \(z\), where \(C\) is a constant. Show that \(f(z)=A z\), where \(A\) is a constant.

Given the complex analytic function \(\Omega(z)=z^{2}\), show that the real part of \(\Omega, \phi(x, y)=\operatorname{Re} \Omega(z)\), satisfies Laplace's equation, \(\nabla_{x, y}^{2} \phi=0\). Let \(z=(1-w) /(1+w)\), where \(w=u+i v\). Show that \(\phi(u, v)=\operatorname{Re} \Omega(w)\) satisfies Laplace's equation \(\nabla_{u, v}^{2} \phi=0\).

Use Morera's Theorem to verify that the following functions are indeed analytic inside a circle of radius \(R\) : (a) \(z^{n}, \quad n \geq 0\) (b) \(e^{z}\) From Morera's Theorem, what can be said about the following functions? (c) \(\frac{\sin z}{z}\) (d) \(\frac{e^{z}}{z}\)

In Cauchy's Integral Formula (Eq. (2.6.1)), take the contour to be a circle of unit radius centered at the origin. Let \(\zeta=e^{i \theta}\) to deduce $$ f(z)=\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{f(\zeta) \zeta}{\zeta-z} d \theta $$ where \(z\) lies inside the circle. Explain why we have $$ 0=\frac{1}{2 \pi} \int_{0}^{2 \pi} \frac{f(\zeta) \zeta}{\zeta-1 / \bar{z}} d \theta $$ and use \(\zeta=1 / \bar{\zeta}\) to show $$ f(z)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\zeta)\left(\frac{\zeta}{\zeta-z} \pm \frac{\bar{z}}{\bar{\zeta}-\bar{z}}\right) d \theta $$ whereupon, using the plus sign $$ f(z)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\zeta) \frac{\left(1-|z|^{2}\right)}{|\zeta-z|^{2}} d \theta $$ (a) Deduce the "Poisson formula" for the real part of \(f(z): u(r, \phi)=\) \(\operatorname{Re} f, z=r e^{i \phi}\) $$ u(r, \phi)=\frac{1}{2 \pi} \int_{0}^{2 \pi} u(\theta) \frac{1-r^{2}}{\left[1-2 r \cos (\phi-\theta)+r^{2}\right]} d \theta $$ where \(u(\theta)=u(1, \theta)\). (b) If we use the minus sign in the formula for \(f(z)\) above, show that $$ f(z)=\frac{1}{2 \pi} \int_{0}^{2 \pi} f(\zeta)\left[\frac{1+r^{2}-2 r e^{i(\theta-\phi)}}{1-2 r \cos (\phi-\theta)+r^{2}}\right] d \theta $$ and by taking the imaginary part $$ v(r, \phi)=C+\frac{1}{\pi} \int_{0}^{2 \pi} u(\theta) \frac{r \sin (\phi-\theta)}{\left[1-2 r \cos (\phi-\theta)+r^{2}\right]} d \theta $$ where \(C=\frac{1}{2 \pi} \int_{0}^{2 \pi} v(1, \theta) d \theta=v(r=0)\). (This last relationship follows from the Cauchy Integral formula at \(z=0-\) see the first equation in this exercise.) (c) Show that $$ \begin{aligned} \frac{2 r \sin (\phi-\theta)}{1-2 r \cos (\phi-\theta)+r^{2}} &=\operatorname{Im}\left[\frac{1-r^{2}+2 i r \sin (\phi-\theta)}{1+r^{2}-2 r \cos (\phi-\theta)}\right] \\ &=\operatorname{Im}\left[\frac{\zeta+z}{\zeta-z}\right] \end{aligned} $$ and therefore the result for \(v(r, \phi)\) from part (b) may be expressed as $$ v(r, \phi)=v(0)+\frac{\operatorname{Im}}{2 \pi} \int_{0}^{2 \pi} u(\theta) \frac{\zeta+z}{\zeta-z} d \theta $$ This example illustrates that prescribing the real part of \(f(z)\) on \(|z|=1\) determines (a) the real part of \(f(z)\) everywhere inside the circle and (b) the imaginary part of \(f(z)\) inside the circle to within a constant. We cannot arbitrarily specify both the real and imaginary parts of an analytic function on \(|z|=1\)