Math 421 Solution of Fall 99 Final

1. Given that the first few terms of the Laurent series for the function $\cot(z)$ around $z=0$ are:
   
   $$\cot z \ \ = \ \ \frac{1}{z} - \frac{z}{3} - \frac{z^3}{45} - \frac{2z^5}{945} - \cdots$$
   
   

   a) Find the principal part at $z=0$ of the function ${\displaystyle f(z) \ = \ \frac{(1+z)\cot z}{z^4}. }$

   Answer: $f(z) = z^{-5} + z^{-4} - \frac{1}{3}z^{-3} - \frac{1}{3}z^{-2} - \frac{1}{45}z^{-1} + \mbox{terms of non-negative degree}.$ The part shown is the principal part.

   b) Find all the singularities of $f(z)$ in the disk $D=\{{\left\vert z\right\vert}<5\}$. Determine the nature of each singularity (isolated, removable, pole of what order, essential).

   Answer: Zero is a pole of order $5$. The function $f(z)$ can be written in the form $\frac{\phi(z)}{\sin(z)}$, where ${\displaystyle \phi(z) \ = \ \frac{(1+z)\cos(z)}{z^4}. }$ Integral multiples of $\pi$ are simple zeroes (of order $1$) of the function $\sin(z)$, of which $-\pi$, $0$, and $\pi$ are in the disk $D$. The function $\phi$ does not vanish at $\pi$ and $-\pi$. Hence, $\pi$ and $-\pi$ are simple poles of $f(z)$.

   c) Find the residue at each isolated singularity in $D$.

   Answer: Using the Laurent serries above, we see that ${\rm Res}_{z=0}f(z)=-\frac{1}{45}$. At the simple pole $\pi$, the residues is

   $$\begin{eqnarray*} {\rm Res}_{z\ = \ \pi}f(z) & = & \lim_{z\rightarrow 0}(z-\pi)f... ...i}{\pi^4}, \\ {\rm Res}_{z=-\pi}f(z) & = & \frac{1-\pi}{\pi^4}. \end{eqnarray*}$$
   
   

2. Compute ${\displaystyle \int_C \frac{\cos z}{e^{iz}-1}dz}$, where $C$ is the circle $\{{\left\vert z\right\vert}=2\}$ (traversed counterclockwise).

   Answer: $g(z):=e^{iz}-1$ vanishes to order $1$ at integer multiples of $2\pi$ (because $g'(2n\pi)=i\neq 0$). The numerator $\cos z$ has value $1$ at $2n\pi$. Thus, $f(z)=\frac{\cos(z)}{e^{iz}-1}$ has a simple pole at $2n\pi$, $n\in {\Bbb Z}$. The only multiple of $2\pi$ enclosed by $C$ is $0$. Using the fact that $0$ is a simple pole, we get
   

   $${\rm Res}_{z=0}f(z) \ = \ \frac{\cos(0)}{{\frac{d}{dz}}_{\mid_{z=0}}(e^{iz}-1)} \ = \ \frac{1}{i} \ = \ -i.$$
   
   
   Cauchy's Theorem yields,
   
   $$\int_Cf(z)dz \ = \ 2\pi i {\rm Res}_{z=0}f(z) \ = \ 2\pi.$$
   
   

3. Compute $I:=\int_C(e^{\sin(z)}+\bar{z})dz$, where $C$ is the circle $\{{\left\vert z\right\vert}=2\}$ (traversed counterclockwise).

   Answer: $e^{\sin(z)}$ is an entire function, so its integral, over any closed contour, is zero (by Cauchy-Goursat's Theorem). Using the parametrization $z=2e^{i\theta}$, we get
   

   $$I = \int_C\bar{z}dz \ = \ \int_0^{2\pi}2e^{-i\theta}2ie^{i\theta}d\theta \ = \ 8\pi i.$$
   
   

4. Compute ${\displaystyle I:=\int_0^{2\pi}\frac{d\theta}{2+\cos(\theta)}. }$

   Answer: Let $z=e^{i\theta}$. Then $\cos(\theta)=\frac{z+\frac{1}{z}}{2}$, $dz=ie^{i\theta}d\theta$ and $d\theta=\frac{dz}{iz}$. The integral gets converted to the contour integral over the unit circle $C$.
   

   $$I=\int_C\frac{1}{2+\left[\frac{z+\frac{1}{z}}{2}\right]}\frac{dz}{iz} \ = \ (-2i)\int_C\frac{dz}{z^2+4z+1}.$$
   
   
   The integrand has poles at $-2\pm\sqrt{3}$. Only $-2+\sqrt{3}$ is enclosed by $C$. We get
   
   $$I \ = \ (2\pi i)(-2i)\cdot {\rm Res}_{z=-2+\sqrt{3}}\frac{1}... ...1}{(-2+\sqrt{3})-(-2-\sqrt{3})} \ = \ \frac{2\pi}{\sqrt{3}}.$$
   
   

5. Compute ${\displaystyle \int_0^\infty \frac{x^2}{1+x^6}dx. }$

   Answer: See the first Example in Section 60 of the text on page 205.

6. a) Find the Laurent series of the function $f(z)=\frac{{\rm Log}z}{z-i}$ around the point $z_0=i$.

   Answer: Let $\sum_{n=0}^\infty a_n (z-i)^n$ be the Taylor series of ${\rm Log}(z)$ centered at $i$. Taylor's Theorem states, in particular, that ${\displaystyle a_n \ = \ \frac{{\rm Log}^{(n)}(i)}{n!}. }$ Now, ${\rm Log}'(z)=\frac{1}{z}$ and ${\displaystyle {\rm Log}^{(n)}(z) = \frac{(-1)^{n+1}(n-1)!}{z^n} }$ for $n\geq 1$. We get, for $n\geq 1$,
   

   $$a_n \ = \ \frac{{\rm Log}^{(n)}(i)}{n!} \ = \ \frac{(-1)^{n+1}}{n i^n} \ = \ \frac{(-1)^{n+1}(-i)^n}{n} \ = \ -\frac{i^n}{n}.$$
   
   
   $a_0={\rm Log}(i)=\frac{\pi i}{2}$. Summerizing, for ${\left\vert z-i\right\vert}<1$ we get
   $$\begin{eqnarray*} {\rm Log}(z) & = & \frac{\pi i}{2}+\sum_{n=1}^\infty -\left(\... ...} +\sum_{k=0}^\infty -\left(\frac{i^{(k+1)}}{k+1}\right)(z-i)^k. \end{eqnarray*}$$
   
   

   b) Find the Taylor series of the function $f(z)=\frac{1}{z^2-3z+2}$ around the point $z_0=0$.

   Answer: Use the partial fraction decomposition
   

   $$\frac{1}{z^2-3z+2} \ = \ \frac{-1}{z-1} + \frac{1}{z-2}.$$
   
   
   Now, use the Taylor series of $\frac{1}{1-w}$ to obtain
   $$\begin{eqnarray*} \frac{-1}{z-1} & = & \frac{1}{1-z} \ = \ \sum_{n=0}^\infty z^... ...m_{n=0}^\infty \left[1-\left(\frac{1}{2}\right)^{n+1}\right]z^n. \end{eqnarray*}$$
   
   

7. Determine whether the following statements are true or false. Justify your answers.

   a) The limit ${\displaystyle \lim_{z\rightarrow 0}\frac{e^{\bar{z}}-1}{z} }$ exists and is equal to $1$.

   Answer: False, $e^{\bar{z}}$ is not analytic at $0$. The above limit is the derivative limit, which does not exists. A direct argument, that the limit doesn't exists, consists of letting $z$ approach $0$ along the $x$ and $y$ axis:

   $$\begin{eqnarray*} \lim_{x\rightarrow 0}\frac{e^x-1}{x} & = & 1, \\ \lim_{iy\rightarrow 0}\frac{e^{-iy}-1}{iy} & = & -1. \end{eqnarray*}$$
   
   

   b) There is a function $f(z)$, analytic in the disk $D=\{{\left\vert z\right\vert}<1\}$, such that
   

   $${\left\vert f(z)\right\vert}^2 \ = \ 4-{\left\vert z\right\vert}^2, \ \ \ \mbox{for all} \ z \ \mbox{in} \ D.$$
   
   

   Answer: False. Such a function $f$ would be a non-constant analytic function, whose absolute value achieves its maximum at the interior point $0$ of $D$. This contradicts the Maximum Modulus Principle.

   c) If $f(z)$ has an isolated singularity at $z_0$ and ${\rm Res}_{z=z_0}(f)=0$, then $z_0$ is a removeable singularity.

   Answer: False. Take $f(z)=\frac{1}{z^2}$ and $z_0=0$ as a counter example.

8. Compute $\cos(\frac{\pi}{2}-i\ln 2).$ Simplify your answer as much as possible.

   Answer:
   

   $$\cos(\frac{\pi}{2}-i\ln 2) \ = \ \frac{e^{i\frac{\pi}{2}+\ln... ...-\ln 2}}{2} \ = \ \frac{2i-\frac{i}{2}}{2} \ = \ \frac{3}{4}i.$$
   
   

9. Prove that ${\displaystyle {\left\vert\int_C e^{iz^2}dz\right\vert}<5, }$ where $C$ is the piece of the circle ${\left\vert z\right\vert}=2$ going from $2$ to $2i$ counter-clockwise.

   Answer: The curve $C$ is parametrized by $z=2e^{i\theta}$, $0\leq \theta \leq \frac{\pi}{2}$. Thus, $z^2=4e^{i2\theta}$ has a positive imaginary part. Consequently, $iz^2$ has a negative real part equal to $-\sin(2\theta)$, where $0\leq \theta \leq \frac{\pi}{2}$, and
   

   $${\left\vert e^{iz^2}\right\vert} \ = \ e^{-\sin(2\theta)} \ < \ 1.$$
   
   
   We conclude, that
   
   $${\left\vert\int_C e^{iz^2}dz\right\vert} \ \leq \ \int_C {... ...q \ 1\cdot \mbox{length}(C) \ = \ \frac{4\pi}{4} \ = \pi < 5.$$
   
   

10. Find an entire function $f(z)$ such that ${\mbox Re}(f) \ = \ 4x^3y-4xy^3-y$.

    Answer: Note the equality
    

    $$(x+iy)^4 \ = \ x^4 + 4ix^3y-6x^2y^2-i4xy^3+y^4.$$
    
    
    Take $f(z)=-iz^4+iz$.