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.
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).
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}}
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 )
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