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