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2.2 exam 1

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Problem 1

Using a well known sum, find a closed for expression for the following series

$$f\left ( z\right ) =1+2z+3z^{2}+4z^{3}+5z^{4}+\cdots$$ Using the ratio test, find for what values of $z$ this series converges.

Solution

Method 1

Assume that the closed form is

$$\left ( 1-z\right ) ^{a}=1+2z+3z^{2}+4z^{3}+5z^{4}+\cdots$$ For some unknown $a$. Now $a$ will be solved for. Using Binomial series definition $\left ( 1-z\right ) ^{a}=1-az+\frac{\left ( a\right ) \left ( a-1\right ) }{2!}z^{2}-\frac{a\left ( a-1\right ) \left ( a-2\right ) }{3!}z^{3}+\cdots$. in the LHS above gives $$1-az+\frac{a\left ( a-1\right ) }{2!}z^{2}-\frac{a\left ( a-1\right ) \left ( a-2\right ) }{3!}z^{3}+\cdots =1+2z+3z^{2}+4z^{3}+5z^{4}+\cdots$$ By comparing coefficients of $z$ in the left side and on the right side shows that $a=-2$ from the coefficient of $z$ term. Verifying this on the coefficient of $z^{2}$ shows it is correct since it gives $\frac{\left ( -2\right ) \left ( -3\right ) }{2}=3$. Therefore $$a=-2$$ The closed form is therefore $$\frac{1}{\left ( 1-z\right ) ^{2}}=1+2z+3z^{2}+4z^{3}+5z^{4}+\cdots$$ Method 2

Starting with Binomial series expansion given by

$$\frac{1}{1-z}=1+z+z^{2}+z^{3}+z^{4}+\cdots$$ Taking derivative w.r.t. $z$ on both sides of the above results in $$\begin{align*} \frac{d}{dz}\left ( \frac{1}{1-z}\right ) & =\frac{d}{dz}\left ( 1+z+z^{2}+z^{3}+z^{4}+\cdots \right ) \\ -\left ( 1-z\right ) ^{-2}\left ( -1\right ) & =0+1+2z+3z^{2}+4z^{3}+\cdots \\ \frac{1}{\left ( 1-z\right ) ^{2}} & =1+2z+3z^{2}+4z^{3}+\cdots \end{align*}$$

Therefore the closed form expression is

$$\frac{1}{\left ( 1-z\right ) ^{2}}=1+2z+3z^{2}+4z^{3}+\cdots$$ Which is the same as method 1.

The series general term of the series is

$$1+2z+3z^{2}+4z^{3}+\cdots =\sum _{n=0}^{\infty }\left ( n+1\right ) z^{n}$$ Applying the ratio test $$\begin{align*} L & =\lim _{n\rightarrow \infty }\left \vert \frac{a_{n+1}}{a_{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+2\right ) z^{n+1}}{\left ( n+1\right ) z^{n}}\right \vert \\ & =\lim _{n\rightarrow \infty }\left \vert \frac{\left ( n+2\right ) z}{n+1}\right \vert \\ & =z\lim _{n\rightarrow \infty }\left \vert \frac{n+2}{n+1}\right \vert \\ & =z\lim _{n\rightarrow \infty }\left \vert \frac{1+\frac{2}{n}}{1+\frac{1}{n}}\right \vert \end{align*}$$

But $\lim _{n\rightarrow \infty }\left \vert \frac{1+\frac{2}{n}}{1+\frac{1}{n}}\right \vert =1$ and the above limit becomes

$$L=z$$ By the ratio test, the series converges when $\left \vert L\right \vert <1$. Therefore $1+2z+3z^{2}+4z^{3}+\cdots$ converges absolutely when $\left \vert z\right \vert <1$.  An absolutely convergent series is also a convergent series. Hence the series converges for $\left \vert z\right \vert <1$.

Problem 2

Find the Laurent series for the function

$$f\left ( z\right ) =\frac{1}{\left ( z^{2}+4\right ) ^{3}}$$ About the isolated singular pole $z=2i$. What is the order of this pole? What is the residue at this pole?

Solution

The poles are at $z^{2}=4$ or $z=\pm 2i$. The expansion of $f\left ( z\right )$ is around the isolated pole at $z=2i$. This pole has order $3$. The region where this expansion is valid is inside a disk centered at $2i$ (but not including the point $z=2i$ itself) and up to the nearest pole which is located at $-2i$. Therefore the disk will have radius $4$.

Let

$$\begin{align*} u & =z-2i\\ z & =u+2i \end{align*}$$

Substituting this expression for $z$ back in $f\left ( z\right )$ gives

$$\begin{align} f\left ( z\right ) & =\frac{1}{\left ( \left ( u+2i\right ) ^{2}+4\right ) ^{3}}\nonumber \\ & =\frac{1}{\left ( u^{2}-4+4ui+4\right ) ^{3}}\nonumber \\ & =\frac{1}{\left ( u^{2}+4ui\right ) ^{3}}\nonumber \\ & =\frac{1}{\left ( u\left ( u+4i\right ) \right ) ^{3}}\nonumber \\ & =\frac{1}{u^{3}}\frac{1}{\left ( u+4i\right ) ^{3}}\nonumber \\ & =\frac{1}{u^{3}}\frac{1}{\left [ 4i\left ( \frac{u}{4i}+1\right ) \right ] ^{3}}\nonumber \\ & =\frac{1}{u^{3}}\frac{1}{\left ( 4i\right ) ^{3}\left ( \frac{u}{4i}+1\right ) ^{3}}\nonumber \\ & =\frac{1}{-i64u^{3}}\frac{1}{\left ( \frac{u}{4i}+1\right ) ^{3}}\nonumber \\ & =\left ( \frac{i}{64u^{3}}\right ) \frac{1}{\left ( \frac{u}{4i}+1\right ) ^{3}} \tag{1} \end{align}$$

Expanding the term $\frac{1}{\left ( 1+\frac{u}{4i}\right ) ^{3}}$ using Binomial series, which is valid for $\left \vert \frac{u}{4i}\right \vert <1$ or $\left \vert u\right \vert <4$ gives

$$\begin{align} \frac{1}{\left ( 1+\frac{u}{4i}\right ) ^{3}} & =1+\left ( -3\right ) \frac{u}{4i}+\frac{\left ( -3\right ) \left ( -4\right ) }{2!}\left ( \frac{u}{4i}\right ) ^{2}+\frac{\left ( -3\right ) \left ( -4\right ) \left ( -5\right ) }{3!}\left ( \frac{u}{4i}\right ) ^{3}+\frac{\left ( -3\right ) \left ( -4\right ) \left ( -5\right ) \left ( -6\right ) }{4!}\left ( \frac{u}{4i}\right ) ^{4}+\cdots \nonumber \\ & =1-3\frac{u}{4i}+\frac{3\cdot 4}{2!}\frac{u^{2}}{16i^{2}}-\frac{3\cdot 4\cdot 5}{3!}\frac{u^{3}}{64i^{3}}+\frac{3\cdot 4\cdot 5\cdot 6}{4!}\frac{u^{4}}{256i^{4}}+\cdots \nonumber \\ & =1+3i\frac{u}{4}-\frac{3\cdot 4}{2!}\frac{u^{2}}{16}-\frac{3\cdot 4\cdot 5}{3!}\frac{u^{3}}{64\left ( -i\right ) }+\frac{3\cdot 4\cdot 5\cdot 6}{4!}\frac{u^{4}}{256}+\cdots \nonumber \\ & =1+3i\frac{u}{4}-\frac{3\cdot 4}{2!}\frac{u^{2}}{16}-i\frac{3\cdot 4\cdot 5}{3!}\frac{u^{3}}{64}+\frac{3\cdot 4\cdot 5\cdot 6}{4!}\frac{u^{4}}{256}+\cdots \tag{2} \end{align}$$

Substituting (2) into (1) and simplifying gives

$$\begin{align*} f\left ( z\right ) & =\left ( \frac{i}{64u^{3}}\right ) \left ( 1+3i\frac{u}{4}-\frac{3\cdot 4}{2!}\frac{u^{2}}{16}-i\frac{3\cdot 4\cdot 5}{3!}\frac{u^{3}}{64}+\frac{3\cdot 4\cdot 5\cdot 6}{4!}\frac{u^{4}}{256}+\cdots \right ) \\ & =\frac{i}{64u^{3}}+\frac{i}{64u^{3}}\left ( 3i\frac{u}{4}\right ) -\frac{i}{64u^{3}}\left ( \frac{3\cdot 4}{2!}\frac{u^{2}}{16}\right ) -\frac{i}{64u^{3}}\left ( i\frac{3\cdot 4\cdot 5}{3!}\frac{u^{3}}{64}\right ) +\frac{i}{64u^{3}}\left ( \frac{3\cdot 4\cdot 5\cdot 6}{4!}\frac{u^{4}}{256}\right ) +\cdots \\ & =\frac{i}{64u^{3}}-\frac{1}{64u^{2}}\frac{3}{4}-\frac{i}{64u}\left ( \frac{3\cdot 4}{2!}\frac{1}{16}\right ) +\frac{1}{64}\left ( \frac{3\cdot 4\cdot 5}{3!}\frac{1}{64}\right ) +\frac{i}{64}\left ( \frac{3\cdot 4\cdot 5\cdot 6}{4!}\frac{u}{256}\right ) +\cdots \\ & =\frac{i}{64u^{3}}-\frac{3}{256u^{2}}-i\frac{3}{512}\frac{1}{u}+\frac{5}{2048}+i\frac{15}{16\,384}u+\cdots \end{align*}$$

Replacing $u$ back by $z-2i$ in the above results in

$$\begin{equation} f\left ( z\right ) =\frac{i}{64}\frac{1}{\left ( z-2i\right ) ^{3}}-\frac{3}{256}\frac{1}{\left ( z-2i\right ) ^{2}}-\frac{3i}{512}\frac{1}{\left ( z-2i\right ) }+\frac{5}{2048}+\frac{15i}{16\,384}\left ( z-2i\right ) +\cdots \tag{3} \end{equation}$$ This expansion is valid for $\left \vert z-2i\right \vert <4$. The above shows that the residue is $$\fbox{$-\frac{3i}{512}$}$$ Which is the coefficient of the $\frac{1}{\left ( z-2i\right ) }$ term in (3).

Problem 3

Use residues to evaluate the following integral

$$I=\int _{0}^{\infty }\frac{dx}{x^{4}+6x^{2}+9}$$ Solution

The integrand is an even function. Therefore the integral $\int _{-\infty }^{\infty }\frac{dx}{x^{4}+6x^{2}+9}$ is evaluated instead and then the required integral $I$ will be half the value obtained. The poles of $\frac{1}{x^{4}+6x^{2}+9}$ are the zeros of the denominator. Factoring the denominator as $\left ( x^{2}+3\right ) \left ( x^{2}+3\right ) =0$, shows the roots are $x=\pm i\sqrt{3}$ from the first factor and $x=\pm i\sqrt{3}$ from the second factor.

Since the upper half plane will be used, the pole located there is $+i\sqrt{3}$ and it is of order two. Now that pole locations are known, the following contour is used to evaluate $\int _{-\infty }^{\infty }\frac{dx}{x^{4}+6x^{2}+9}$ as shown in the plot below

$$\begin{align}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz & =\lim _{R\rightarrow \infty }\int _{C}f\left ( z\right ) dz+\lim _{R\rightarrow \infty }\int _{-R}^{+R}f\left ( x\right ) dx\nonumber \\ & =\lim _{R\rightarrow \infty }\int _{C}\frac{dz}{z^{4}+6z^{2}+9}+\lim _{R\rightarrow \infty }\int _{-R}^{+R}\frac{dx}{x^{4}+6x^{2}+9}dx \tag{2} \end{align}$$

Where the integral $\int _{-R}^{+R}$ is Cauchy principal integral. Since the contour $C$ is closed and because $f\left ( z\right )$ is analytic on and inside $C$ except for the isolated singularity inside at $z=i\sqrt{3}$, then by Cauchy integral formula ${\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=2\pi i\sum \operatorname{Residue}$. Where the sum of residues is over all poles inside $C$. Therefore (2) can becomes

$$\begin{equation} \int _{-\infty }^{+\infty }\frac{dx}{x^{4}+6x^{2}+9}dx=2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C}f\left ( z\right ) dz \tag{3} \end{equation}$$ But $$\begin{align} \left \vert \int _{C}f\left ( z\right ) dz\right \vert _{\max } & \leq ML\nonumber \\ & =\left \vert f\left ( z\right ) \right \vert _{\max }\pi R \tag{4} \end{align}$$

Using

$$\left \vert f\left ( z\right ) \right \vert _{\max }\leq \frac{1}{\left \vert z^{2}+3\right \vert _{\min }\left \vert z^{2}+3\right \vert _{\min }}$$ By inverse triangle inequality $\left \vert z^{2}+3\right \vert \geq \left \vert z\right \vert ^{2}-3$. But $\left \vert z\right \vert =R$ on $C$, therefore $\left \vert z^{2}+3\right \vert \geq R^{2}-3$ and the above can now be written as $$\left \vert f\left ( z\right ) \right \vert _{\max }\leq \frac{1}{\left ( R^{2}-3\right ) \left ( R^{2}-3\right ) }$$ Using the above in (4) gives $$\begin{align*} \left \vert \int _{C}f\left ( z\right ) dz\right \vert _{\max } & \leq \frac{\pi R}{\left ( R^{2}-3\right ) \left ( R^{2}-3\right ) }\\ & =\frac{\pi R}{R^{4}-6R^{2}+9}\\ & =\frac{\frac{\pi }{R}}{R^{2}-6+\frac{9}{R^{2}}} \end{align*}$$

In the limit as $R\rightarrow \infty$ then $\left \vert \int _{C}f\left ( z\right ) dz\right \vert _{\max }\rightarrow 0$. Using this result in (3) it simplifies to

$$\begin{equation} \int _{-\infty }^{+\infty }\frac{dx}{x^{4}+6x^{2}+9}dx=2\pi i\sum \operatorname{Residue} \tag{5} \end{equation}$$ What is left now is to determine the residue at pole $z_{0}=i\sqrt{3}$ which is of order $2$. This is done using $$\operatorname{Residue}\left ( z_{0}\right ) =\lim _{z\rightarrow z_{0}}\frac{d}{dz}\left ( \left ( z-z_{0}\right ) ^{2}f\left ( z\right ) \right )$$ But $z_{0}=i\sqrt{3}$ and the above becomes $$\begin{align*} \operatorname{Residue}\left ( i\sqrt{3}\right ) & =\lim _{z\rightarrow i\sqrt{3}}\frac{d}{dz}\left ( \left ( z-i\sqrt{3}\right ) ^{2}\frac{1}{\left ( z-i\sqrt{3}\right ) ^{2}\left ( z+i\sqrt{3}\right ) ^{2}}\right ) \\ & =\lim _{z\rightarrow i\sqrt{3}}\frac{d}{dz}\frac{1}{\left ( z+i\sqrt{3}\right ) ^{2}}\\ & =\lim _{z\rightarrow i\sqrt{3}}\frac{-2}{\left ( z+i\sqrt{3}\right ) ^{3}}\\ & =\frac{-2}{\left ( i\sqrt{3}+i\sqrt{3}\right ) ^{3}}\\ & =\frac{-2}{\left ( 2i\sqrt{3}\right ) ^{3}}\\ & =\frac{-2}{-\left ( 8\right ) \left ( 3\right ) i\sqrt{3}}\\ & =\frac{1}{12i\sqrt{3}} \end{align*}$$

Using the above value of the residue in (5) gives

$$\begin{align*} \int _{-\infty }^{+\infty }\frac{dx}{x^{4}+6x^{2}+9}dx & =2\pi i\left ( \frac{1}{12i\sqrt{3}}\right ) \\ & =\frac{\pi }{6\sqrt{3}} \end{align*}$$

Therefore the integral $\int _{0}^{\infty }\frac{dx}{x^{4}+6x^{3}+9}$ is half of the above result which is

$$\int _{0}^{\infty }\frac{dx}{x^{4}+6x^{2}+9}=\frac{\pi }{12\sqrt{3}}$$

Problem 4

Find two approximations for the integral $x>0$

$$I\left ( x\right ) =\frac{1}{2\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{x\cos ^{2}\theta }d\theta$$ One for small $x$ (keeping up to linear order in $x$) and one for large values of $x$ (keeping only the leading order term).

Solution

The integrand has the form $e^{z}$. This has a known Taylor series expansion around zero given by

$$e^{z}=1+z+\frac{z^{2}}{2!}+\cdots$$ Replacing $z$ by $x\cos ^{2}\theta$ in the above gives $$e^{x\cos ^{2}\theta }=1+x\cos ^{2}\theta +\frac{\left ( x\cos ^{2}\theta \right ) ^{2}}{2}+\cdots$$ The problem is asking to keep linear terms in $x$. Therefore $$e^{x\cos ^{2}\theta }\approx 1+x\cos ^{2}\theta$$ Replacing the integrand in the original integral by the above approximation gives $$\begin{align*} I\left ( x\right ) & \approx \frac{1}{2\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\left ( 1+x\cos ^{2}\theta \right ) d\theta \\ & \approx \frac{1}{2\pi }\left ( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}d\theta +x\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\cos ^{2}\theta d\theta \right ) \\ & \approx \frac{1}{2\pi }\left ( \int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}d\theta +x\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{1}{2}+\frac{1}{2}\cos 2\theta d\theta \right ) \\ & \approx \frac{1}{2\pi }\left ( \pi +x\left ( \frac{1}{2}\theta +\frac{1}{2}\frac{\sin 2\theta }{2}\right ) _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\right ) \\ & \approx \frac{1}{2\pi }\left ( \pi +\frac{x}{4}\left ( 2\theta +\sin 2\theta \right ) _{-\frac{\pi }{2}}^{\frac{\pi }{2}}\right ) \\ & \approx \frac{1}{2\pi }\left ( \pi +\frac{x}{4}\left ( 2\pi +0\right ) \right ) \\ & \approx \frac{1}{2\pi }\left ( \pi +\frac{x}{2}\pi \right ) \\ & \approx \frac{1}{2}\left ( 1+\frac{x}{2}\right ) \end{align*}$$

For large value of $x$, The integrand is written as $e^{f\left ( \theta \right ) }$ where $f\left ( \theta \right ) =x\cos ^{2}\theta$. The value of $\theta$ where $f\left ( \theta \right )$ is maximum is first found. Then solving for $\theta$ in

$$\begin{align*} \frac{d}{d\theta }x\cos ^{2}\theta & =0\\ -2x\cos \theta \sin \theta & =0 \end{align*}$$

Hence solving for $\theta$ in

$$\cos \theta \sin \theta =0$$ There are two solutions to this. Either $\theta =\frac{\pi }{2}$ or $\theta =0$. To find which is the correct choice, the sign of $\frac{d^{2}}{d\theta ^{2}}f\left ( \theta \right )$ is checked for each choice. $$\begin{align} \frac{d^{2}}{d\theta ^{2}}x\cos ^{2}\theta & =\frac{d}{d\theta }\left ( -2x\cos \theta \sin \theta \right ) \nonumber \\ & =-2x\frac{d}{d\theta }\left ( \cos \theta \sin \theta \right ) \nonumber \\ & =-2x\left ( -\sin \theta \sin \theta +\cos \theta \cos \theta \right ) \nonumber \\ & =-2x\left ( -\sin ^{2}\theta +\cos ^{2}\theta \right ) \tag{1} \end{align}$$

Substituting $\theta =\frac{\pi }{2}$ in (1) and using $\cos \left ( \frac{\pi }{2}\right ) =0$ and $\sin \left ( \frac{\pi }{2}\right ) =1$ gives

$$\begin{align*} \left . \frac{d^{2}}{d\theta ^{2}}x\cos ^{2}\theta \right \vert _{\theta =\frac{\pi }{2}} & =-2x\left ( -1\right ) \\ & =2x \end{align*}$$

Since the problem says that $x>0$ then $\left . \frac{d^{2}}{d\theta ^{2}}x\cos ^{2}\theta \right \vert _{\theta =\frac{\pi }{2}}>0$. Therefore this is a minimum. Using the second choice $\theta =0$, then (1) becomes (after using $\cos \left ( 0\right ) =1$ and $\sin \left ( 0\right ) =0$)

$$\left . \frac{d^{2}}{d\theta ^{2}}x\cos ^{2}\theta \right \vert _{\theta =0}=-2x$$ And because $x>0$ then $\left . \frac{d^{2}}{d\theta ^{2}}x\cos ^{2}\theta \right \vert _{\theta =0}<0$. Therefore the integrand is maximum at $$\theta _{peak}=0$$ Now that peak $\theta$ is  found, then $f\left ( \theta \right )$ is expanded in Taylor series around $\theta _{peak}=0.$ Since $f\left ( \theta \right ) =x\cos ^{2}\theta$, then $$f\left ( \theta _{peak}\right ) =x$$ And $f^{\prime }\left ( \theta \right ) =-2x\cos \theta \sin \theta$. At $\theta _{peak}$ this becomes $f^{\prime }\left ( \theta _{peak}\right ) =0$. The next term is the quadratic one, given by $$\begin{align*} f^{\prime \prime }\left ( \theta \right ) & =-2x\frac{d}{d\theta }\left ( \cos \theta \sin \theta \right ) \\ & =-2x\left ( -\sin ^{2}\theta +\cos ^{2}\theta \right ) \end{align*}$$

Evaluating the above at $\theta _{peak}=0$ gives

$$f^{\prime \prime }\left ( \theta _{peak}\right ) =-2x$$ The problem says to keep leading term, so no need for more terms. Therefore the Taylor series expansion of $f\left ( \theta \right ) =x\cos ^{2}\theta$ around $\theta =\theta _{peak}$ is $$\begin{align*} x\cos ^{2}\theta & \approx f\left ( \theta _{peak}\right ) +f^{\prime }\left ( \theta _{peak}\right ) \theta +\frac{1}{2!}f^{\prime \prime }\left ( \theta _{peak}\right ) \theta ^{2}\\ & =x+0-\frac{2x}{2!}\theta ^{2}\\ & =x-x\theta ^{2}\\ & =x\left ( 1-\theta ^{2}\right ) \end{align*}$$

The integral now becomes

$$\begin{align*} I\left ( x\right ) & =\frac{1}{2\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{x\left ( 1-\theta ^{2}\right ) }d\theta \\ & \approx \frac{1}{2\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{x}e^{-x\theta ^{2}}d\theta \\ & =\frac{1}{2\pi }\left ( e^{x}\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{-x\theta ^{2}}d\theta \right ) \end{align*}$$

Comparing $\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{-x\theta ^{2}}d\theta$ to the Gaussian integral $\int _{-\infty }^{\infty }e^{-a\theta ^{2}}d\theta =\sqrt{\frac{\pi }{x}}$, then the above can be approximated as

$$I\left ( x\right ) =\frac{e^{x}}{2\pi }\sqrt{\frac{\pi }{x}}$$ Summary of result

Small $x$ approximation	$\frac{1}{2}\left ( 1+\frac{x}{2}\right )$
Large $x$ approximation	$\frac{e^{x}}{2\pi }\sqrt{\frac{\pi }{x}}$

Note that using the computer, the exact solution is

$$\frac{1}{2\pi }\int _{-\frac{\pi }{2}}^{\frac{\pi }{2}}e^{x\cos ^{2}\theta }d\theta =\frac{1}{2}e^{\frac{x}{2}}\operatorname{BesselI}\left ( 0,\frac{x}{2}\right )$$

Problem 5

Use the Cauchy-Riemann equations to determine where the function

$$f\left ( z\right ) =z+\overline{z^{2}}$$ Is analytic. Evaluate ${\displaystyle \oint \limits _{C}} f\left ( z\right ) dz$ where contour $C$ is on the unit circle $\left \vert z\right \vert =1$ in a counterclockwise sense.

Solution

Using $z=x+iy$, the function $f\left ( z\right )$ becomes

$$\begin{align*} f\left ( z\right ) & =x+iy+\overline{\left ( x+iy\right ) ^{2}}\\ & =x+iy+\overline{\left ( x^{2}-y^{2}+2ixy\right ) }\\ & =x+iy+\left ( x^{2}-y^{2}-2ixy\right ) \\ & =\left ( x+x^{2}-y^{2}\right ) +i\left ( y-2xy\right ) \end{align*}$$

Writing $f\left ( z\right ) =u+iv$, and comparing this to the above result shows that

$$\begin{align} u & =x+x^{2}-y^{2}\nonumber \\ v & =y-2xy \tag{1} \end{align}$$

Cauchy-Riemann are given by

$$\begin{align*} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\\ \frac{-\partial u}{\partial y} & =\frac{\partial v}{\partial x} \end{align*}$$

Using result in (1), Cauchy-Riemann are checked to see if they are satisfied or not. The first equation above results in

$$\begin{align*} \frac{\partial u}{\partial x} & =1+2x\\ \frac{\partial v}{\partial y} & =1-2x \end{align*}$$

Therefore $\frac{\partial u}{\partial x}\neq \frac{\partial v}{\partial y}$. This shows that $f\left ( z\right )$ is not analytic for all $x,y$.

Since $f\left ( z\right )$ is not analytic, Cauchy integral formula can not be used. Instead this can be integrated using parameterization. Let $z=e^{i\theta }$ (No need to use $re^{i\theta }$ since $r=1$ in this case because it is the unit circle). The function $f\left ( z\right )$ becomes

$$\begin{align*} f\left ( z\right ) & =e^{i\theta }+\overline{\left ( e^{i\theta }\right ) ^{2}}\\ & =e^{i\theta }+\overline{e^{2i\theta }}\\ & =e^{i\theta }+e^{-2i\theta } \end{align*}$$

And because $z=e^{i\theta }$ then $dz=d\theta e^{i\theta }$. The integral now becomes

$$\begin{align*}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz & =\int _{0}^{2\pi }\left ( e^{i\theta }+e^{-2i\theta }\right ) e^{i\theta }d\theta \\ & =\int _{0}^{2\pi }\left ( e^{2i\theta }+e^{-i\theta }\right ) d\theta \\ & =\left [ \frac{e^{2i\theta }}{2i}\right ] _{0}^{2\pi }+\left [ \frac{e^{-i\theta }}{-i}\right ] _{0}^{2\pi }\\ & =\frac{1}{2i}\left [ \cos 2\theta +i\sin 2\theta \right ] _{0}^{2\pi }+i\left [ \cos \theta -i\sin \theta \right ] _{0}^{2\pi }\\ & =\frac{1}{2i}\left [ \left ( \cos 4\pi +i\sin 4\pi \right ) -\left ( \cos 0+i\sin 0\right ) \right ] +i\left [ \left ( \cos 2\pi -i\sin 2\pi \right ) -\left ( \cos 0-i\sin 0\right ) \right ] \\ & =\frac{1}{2i}\left [ 1-1\right ] +i\left [ 1-1\right ] \end{align*}$$

Hence

$${\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=0$$