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3.2 HW 2

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3.2.1 Problem 1

Find all possible values for (put into $x+iy$ form)

1.

$\log \left ( 1+\sqrt{3}i\right )$

2.

$\left ( 1+\sqrt{3}i\right ) ^{2i}$

Answer

Part 1

Let $z=x+iy$, where here $x=1,y=3$, then $\left \vert z\right \vert =\sqrt{x^{2}+y^{2}}=\sqrt{1+3}=2$ and $\arg \left ( z\right ) =\theta _{0}=\arctan \left ( \frac{y}{x}\right ) =\arctan \left ( \frac{\sqrt{3}}{1}\right ) =\frac{\pi }{6}=60^{0}$.  The function $\log \left ( z\right )$ is infinitely multi-valued, given by

$$\begin{equation} \log \left ( z\right ) =\ln \left \vert z\right \vert +i\left ( \theta _{0}+2n\pi \right ) \qquad n=0,\pm 1,\pm 2,\cdots \tag{1} \end{equation}$$ Where $\theta _{0}$ is the principal argument, which is $60^{0}$ in this example, which is when $n=0$. This is done to make $\log \left ( z\right )$ single valued. This makes the argument of $z$ restricted to $-\pi <\theta _{0}<\pi$. This makes the negative real axis the branch cut, including the origin. To find all values, we simply use (1) for all possible $n$ values other than $n=0$. Each different $n$ values gives different branch cut. This gives, where $\ln \left \vert z\right \vert =\ln \left ( 2\right )$ in all cases, the following $$\begin{align*} \log \left ( z\right ) & =\ln \left ( 2\right ) +i\left ( \frac{\pi }{3}\right ) \qquad n=0\qquad \\ & =\ln \left ( 2\right ) +i\left ( \frac{\pi }{3}+2\pi \right ) \qquad n=1\qquad \\ & =\ln \left ( 2\right ) +i\left ( \frac{\pi }{3}-2\pi \right ) \qquad n=-1\qquad \\ & =\ln \left ( 2\right ) +i\left ( \frac{\pi }{3}+4\pi \right ) \qquad n=2\qquad \\ & =\ln \left ( 2\right ) +i\left ( \frac{\pi }{3}-4\pi \right ) \qquad n=-2\qquad \\ & \vdots \end{align*}$$

Or

$$\begin{align*} \log \left ( z\right ) & =0.693+1.047i\\ & =0.693+7.330i\\ & =0.693-5.236i\\ & =0.693+13.614i\\ & =0.693-11.519i\\ & \vdots \end{align*}$$

These are in $\left ( x+iy\right )$ form. There are infinite number of values. Picking a specific branch cuts (i.e. specific $n$ value), picks one of these values. The principal value is one associated with $n=0$.

Part 2

Let $z=1+i\sqrt{3}$, hence

$$\begin{align*} f\left ( z\right ) & =z^{2i}\\ & =\exp \left ( 2i\log \left ( z\right ) \right ) \\ & =\exp \left ( 2i\left ( \ln \left \vert z\right \vert +i\left ( \theta _{0}+2n\pi \right ) \right ) \right ) \qquad n=0,\pm 1,\pm 2,\cdots \end{align*}$$

Where in this example, as in first part, $\ln \left \vert z\right \vert =\ln \left ( 2\right ) =0.693$ and principal argument is $\theta _{0}=\frac{\pi }{3}=60^{0}$. Hence the above becomes

$$\begin{align*} f\left ( z\right ) & =\exp \left ( 2i\left ( \ln \left ( 2\right ) +i\left ( \frac{\pi }{3}+2n\pi \right ) \right ) \right ) \\ & =\exp \left ( 2i\ln \left ( 2\right ) -\left ( \frac{2\pi }{3}+4n\pi \right ) \right ) \\ & =\exp \left ( i\ln 4-\left ( \frac{2\pi }{3}+4n\pi \right ) \right ) \\ & =\exp \left ( i\ln 4\right ) \exp \left ( -\left ( \frac{2\pi }{3}+4n\pi \right ) \right ) \\ & =e^{-\left ( \frac{2\pi }{3}+4n\pi \right ) }\left ( \cos \left ( \ln 4\right ) +i\sin \left ( \ln 4\right ) \right ) \\ & =e^{-\left ( \frac{2\pi }{3}+4n\pi \right ) }\cos \left ( \ln 4\right ) +ie^{-\left ( \frac{2\pi }{3}+4n\pi \right ) }\sin \left ( \ln 4\right ) \end{align*}$$

Which is now in the form of $x+iy$. First few values are

$$\begin{align*} f\left ( z\right ) & =e^{-\left ( \frac{2\pi }{3}\right ) }\cos \left ( \ln 4\right ) +ie^{-\left ( \frac{2\pi }{3}\right ) }\sin \left ( \ln 4\right ) \qquad n=0\\ & =e^{-\left ( \frac{2\pi }{3}+4\pi \right ) }\cos \left ( \ln 4\right ) +ie^{-\left ( \frac{2\pi }{3}+4\pi \right ) }\sin \left ( \ln 4\right ) \qquad n=1\\ & =e^{-\left ( \frac{2\pi }{3}-4\pi \right ) }\cos \left ( \ln 4\right ) +ie^{-\left ( \frac{2\pi }{3}-4\pi \right ) }\sin \left ( \ln 4\right ) \qquad n=-1\\ & =e^{-\left ( \frac{2\pi }{3}+8\pi \right ) }\cos \left ( \ln 4\right ) +ie^{-\left ( \frac{2\pi }{3}+8\pi \right ) }\sin \left ( \ln 4\right ) \qquad n=2\\ & =e^{-\left ( \frac{2\pi }{3}-8\pi \right ) }\cos \left ( \ln 4\right ) +ie^{-\left ( \frac{2\pi }{3}-8\pi \right ) }\sin \left ( \ln 4\right ) \qquad n=-2\\ & \vdots \end{align*}$$

Or

$$\begin{align*} f\left ( z\right ) & =0.0226+i0.121\,\\ & =7.878\,\times 10^{-8}+i4.222\times 10^{-7}\\ & =6478+i34713\\ & =2.748\times 10^{-13}+i1.472\times 10^{-12}\\ & =1.858\times 10^{9}+i9.954\times 10^{9}\\ & \vdots \end{align*}$$

3.2.2 Problem 2

Given that $u\left ( x,y\right ) =3x^{2}y-y^{3}$ find $v\left ( x,y\right )$ such that $f\left ( z\right )$ is analytic. Do the same for $u\left ( x,y\right ) =\frac{y}{x^{2}+y^{2}}$

Solution

Part (1)

$u\left ( x,y\right ) =3x^{2}y-y^{3}.$ The function $f\left ( z\right )$ is analytic if it satisfies Cauchy-Riemann equations

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

Applying the first equation gives

$$6xy=\frac{\partial v}{\partial y}$$ Hence, solving for $v$ by integrating, gives $$\begin{equation} v\left ( x,y\right ) =3xy^{2}+f\left ( x\right ) \tag{3} \end{equation}$$ Is the solution to (3) where $f\left ( x\right )$ is the constant of integration since it is a partial differential equation. We now use equation (2) to find $f\left ( x\right )$. From (2) $$\begin{align*} -\left ( 3x^{2}-3y^{2}\right ) & =\frac{\partial v}{\partial x}\\ -3x^{2}+3y^{2} & =\frac{\partial v}{\partial x} \end{align*}$$

But (3) gives $\frac{\partial v}{\partial x}=3y^{2}+f^{\prime }\left ( x\right )$, hence the above becomes

$$\begin{align*} -3x^{2}+3y^{2} & =3y^{2}+f^{\prime }\left ( x\right ) \\ f^{\prime }\left ( x\right ) & =-3x^{2}+3y^{2}-3y^{2}\\ & =-3x^{2} \end{align*}$$

Integrating gives

$$\begin{align*} f\left ( x\right ) & =\int -3x^{2}dx\\ & =-x^{3}+C \end{align*}$$

Therefore, (3) becomes

$$v\left ( x,y\right ) =3xy^{2}+f\left ( x\right )$$ Or $$\fbox{$v\left ( x,y\right ) =3xy^2-x^3+C$}$$ Where $C$ is arbitrary constant. To verify, we apply CR again. Equation (1) now gives $$\begin{align*} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\\ 6xy & =6yx \end{align*}$$

Verified. Equation (2) gives

$$\begin{align*} -\frac{\partial u}{\partial y} & =\frac{\partial v}{\partial x}\\ -3x^{2}+3y^{2} & =-3x^{2}+3y^{2} \end{align*}$$

Verified.

Part (2)

$u\left ( x,y\right ) =\frac{y}{x^{2}+y^{2}}.$ The function $f\left ( z\right )$ is analytic if it satisfies Cauchy-Riemann equations

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

Applying the first equation gives

$$-\frac{2xy}{\left ( x^{2}+y^{2}\right ) ^{2}}=\frac{\partial v}{\partial y}$$ Hence, solving for $v$ by integrating, gives $$\begin{align} v & =-2x\int \frac{y}{\left ( x^{2}+y^{2}\right ) ^{2}}dy\nonumber \\ & =\frac{x}{x^{2}+y^{2}}+f\left ( x\right ) \tag{3} \end{align}$$

Is the solution to (3) where $f\left ( x\right )$ is the constant of integration since it is a partial differential equation. equation (2) gives

$$-\frac{1}{x^{2}+y^{2}}+\frac{2y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}=\frac{\partial v}{\partial x}$$ But (3) gives $\frac{\partial v}{\partial x}=\frac{1}{x^{2}+y^{2}}-\frac{2x^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}+f^{\prime }\left ( x\right )$, hence the above becomes $$\begin{align*} -\frac{1}{x^{2}+y^{2}}+\frac{2y^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}} & =\frac{1}{x^{2}+y^{2}}-\frac{2x^{2}}{\left ( x^{2}+y^{2}\right ) ^{2}}+f^{\prime }\left ( x\right ) \\ f^{\prime }\left ( x\right ) & =-\frac{2}{x^{2}+y^{2}}+\frac{2\left ( y^{2}+x^{2}\right ) }{\left ( x^{2}+y^{2}\right ) ^{2}}\\ & =-\frac{2}{x^{2}+y^{2}}+\frac{2}{\left ( x^{2}+y^{2}\right ) }\\ & =0 \end{align*}$$

Hence

$$f\left ( x\right ) =C$$ where $C$ is arbitrary constant. Therefore, (3) becomes $$\fbox{$v\left ( x,y\right ) =\frac{x}{x^2+y^2}+C$}$$ To verify, CR is applied again. Equation (1) now gives $$\begin{align*} \frac{\partial u}{\partial x} & =\frac{\partial v}{\partial y}\\ \frac{-2xy}{\left ( x^{2}+y^{2}\right ) ^{2}} & =\frac{-2xy}{\left ( x^{2}+y^{2}\right ) ^{2}} \end{align*}$$

Hence verified. Equation (2) gives

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

Verified.

3.2.3 Problem 3

Evaluate the integral (i) ${\displaystyle \oint \limits _{C}} \left \vert z\right \vert ^{2}dz$ and (ii) ${\displaystyle \oint \limits _{C}} \frac{1}{z^{2}}dz$ along two contours. These contours are

1.

Line segment with initial point $1$ and fixed point $i$

2.

Arc of unit circle with $\operatorname{Im}\left ( z\right ) \geq 0$ with initial point $1$ and final point $i$

Solution

Part (1)

Figure 3.1:Integration path

First integral We start by finding the parameterization. For line segments that starts at $\left ( x_{0},y_{0}\right )$ and ends at $\left ( x_{1},y_{1}\right )$, the parametrization is given by

$$\begin{align*} x\left ( t\right ) & =\left ( 1-t\right ) x_{0}+tx_{1}\\ y\left ( t\right ) & =\left ( 1-t\right ) y_{0}+ty_{1} \end{align*}$$

For $0\leq t\leq 1$. Hence for $z=x+iy$, it becomes $z\left ( t\right ) =x\left ( t\right ) +iy\left ( t\right )$. In this case, $x_{0}=1,y_{0}=0,x_{1}=0,y_{1}=1$, therefore

$$\begin{align*} x\left ( t\right ) & =\left ( 1-t\right ) \\ y\left ( t\right ) & =t \end{align*}$$

Using these, $z\left ( t\right )$ is found from

$$\begin{align*} z\left ( t\right ) & =x\left ( t\right ) +iy\left ( t\right ) \\ & =\left ( 1-t\right ) +it \end{align*}$$

And

$$z^{\prime }\left ( t\right ) =-1+i$$ Since $\left \vert z\right \vert ^{2}=x^{2}+y^{2}$, then in terms of $t$ it becomes $$\left \vert z\left ( t\right ) \right \vert ^{2}=\left ( 1-t\right ) ^{2}+t^{2}$$ Hence the line integral now becomes $$\begin{align*} \int _{C}\left \vert z\right \vert ^{2}dz & =\int _{0}^{1}\left \vert z\left ( t\right ) \right \vert ^{2}z^{\prime }\left ( t\right ) dt\\ & =\int _{0}^{1}\left ( \left ( 1-t\right ) ^{2}+t^{2}\right ) \left ( -1+i\right ) \,dt\\ & =\left ( -1+i\right ) \int _{0}^{1}\left ( 1-t\right ) ^{2}+t^{2}dt\\ & =\left ( -1+i\right ) \int _{0}^{1}1+t^{2}-2t+t^{2}dt\\ & =\left ( -1+i\right ) \int _{0}^{1}1+2t^{2}-2t\ dt\\ & =\left ( -1+i\right ) \left ( \int _{0}^{1}dt+\int _{0}^{1}2t^{2}dt-\int _{0}^{1}2t\ dt\right ) \\ & =\left ( -1+i\right ) \left ( \left ( t\right ) _{0}^{1}+2\left ( \frac{t^{3}}{3}\right ) _{0}^{1}-2\left ( \frac{t^{2}}{2}\right ) _{0}^{1}\right ) \\ & =\left ( -1+i\right ) \left ( 1+\frac{2}{3}-2\left ( \frac{1}{2}\right ) \right ) \end{align*}$$

Hence

$$\fbox{$\int _C\left \vert z\right \vert ^2dz=\frac{2}{3}\left ( i-1\right ) $}$$ second integral

Using the same parameterization above. But here the integrand is

$$\frac{1}{z^{2}}=\frac{1}{\left ( \left ( 1-t\right ) +it\right ) ^{2}}$$ Hence the integral becomes $$\begin{align*} \int _{C}\frac{1}{z^{2}}dz & =\int _{0}^{1}\frac{1}{\left ( \left ( 1-t\right ) +it\right ) ^{2}}z^{\prime }\left ( t\right ) dt\\ & =\left ( i-1\right ) \int _{0}^{1}\frac{1}{\left ( \left ( 1-t\right ) +it\right ) ^{2}}dt\\ & =\left ( i-1\right ) \left ( -i\right ) \end{align*}$$

Hence

$$\fbox{$\int _C\frac{1}{z^2}dz=1+i$}$$

Part (2)

Figure 3.2:Integration path

First integral Let $z=re^{i\theta }$ then $\frac{dz}{d\theta }=rie^{i\theta }$. When $z=1$ then $\theta =0$. When $z=i$ then $\theta =\frac{\pi }{2}$, hence we can parameterize the contour integral using $\theta$ and it becomes

$$\begin{align*} \int _{C}\left \vert z\right \vert ^{2}dz & =\int _{0}^{\frac{\pi }{2}}r^{2}\left ( rie^{i\theta }\right ) d\theta \\ & =ir^{3}\int _{0}^{\frac{\pi }{2}}e^{i\theta }d\theta \\ & =ir^{3}\left [ \frac{e^{i\theta }}{i}\right ] _{0}^{\frac{\pi }{2}}\\ & =r^{3}\left [ e^{i\theta }\right ] _{0}^{\frac{\pi }{2}}\\ & =r^{3}\left [ e^{i\frac{\pi }{2}}-e^{0}\right ] \\ & =r^{3}\left [ i-1\right ] \end{align*}$$

But $r=1$, therefore the above becomes

$$\fbox{$\int _C\left \vert z\right \vert ^2dz=i-1$}$$ second integral

Using the same parameterization above. But here the integrand now

$$\frac{1}{z^{2}}=\frac{1}{r^{2}e^{i2\theta }}$$ Therefore $$\begin{align*} \int _{C}\frac{1}{z^{2}}dz & =\int _{0}^{\frac{\pi }{2}}\frac{1}{r^{2}e^{i2\theta }}\left ( rie^{i\theta }\right ) d\theta \\ & =\frac{i}{r}\int _{0}^{\frac{\pi }{2}}e^{-i\theta }d\theta \\ & =\frac{i}{r}\left ( \frac{e^{-i\theta }}{-i}\right ) _{0}^{\frac{\pi }{2}}\\ & =\frac{-1}{r}\left ( e^{-i\theta }\right ) _{0}^{\frac{\pi }{2}}\\ & =\frac{-1}{r}\left ( e^{-i\frac{\pi }{2}}-1\right ) \\ & =\frac{-1}{r}\left ( -i-1\right ) \end{align*}$$

But $r=1$, hence

$$\fbox{$\int _C\frac{1}{z^2}dz=1+i$}$$

3.2.4 Problem 4

Use the Cauchy integral formula

$$f\left ( z_{0}\right ) =\frac{1}{2\pi i}{\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{z-z_{0}}dz$$

To evaluate

$${\displaystyle \oint \limits _{C}} \frac{1}{\left ( z+1\right ) \left ( z+2\right ) }dz$$ Where $C$ is the circular contour $\left \vert z+1\right \vert =R$ with $R<1$. Note that if $R>1$ then a different result is found. Why can’t the Cauchy integral formula above be used for $R>1$?

Solution

The disk $\left \vert z+1\right \vert =R$ is centered at $z=-1$ with $R<1$. The function

$$g\left ( z\right ) =\frac{1}{\left ( z+1\right ) \left ( z+2\right ) }$$ has pole at $z=-1$ and at $z=-2$.

Figure 3.3:Showing location of pole

In the Cauchy integral formula, the function $f\left ( z\right )$ is analytic on $C$ and inside $C$. Hence, to use Cauchy integral formula, we need to convert $g\left ( z\right ) =\frac{1}{\left ( z+1\right ) \left ( z+2\right ) }$ to look like $\frac{f\left ( z\right ) }{z-z_{0}}$ where $f\left ( z\right )$ is analytic inside $C$. This is done as follows

$$\begin{align*} \frac{1}{\left ( z+1\right ) \left ( z+2\right ) } & =\frac{\frac{1}{\left ( z+2\right ) }}{z-\left ( -1\right ) }\\ & =\frac{f\left ( z\right ) }{z-\left ( -1\right ) } \end{align*}$$

Where now $f\left ( z\right ) =\frac{1}{\left ( z+2\right ) }$. This has pole at $z=-2$. Since this pole is outside $C$ then $f\left ( z\right )$ is analytic on and inside $C$ and can be used for the purpose of using Cauchy integral formula, which now can be written as

$$\begin{align*}{\displaystyle \oint \limits _{C}} \frac{1}{\left ( z+1\right ) \left ( z+2\right ) }dz & ={\displaystyle \oint \limits _{C}} \frac{\frac{1}{\left ( z+2\right ) }}{z-\left ( -1\right ) }dz\\ & ={\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{z-\left ( -1\right ) }dz\\ & =\left ( 2\pi i\right ) f\left ( -1\right ) \end{align*}$$

Therefore, we just need to evaluate $f\left ( -1\right )$ which is seen as $1$. Hence

$$\begin{equation} \fbox{${\displaystyle \oint \limits _C} \frac{1}{\left ( z+1\right ) \left ( z+2\right ) }dz=2\pi i$} \tag{1} \end{equation}$$ To verify, we can solve this again using the residue theorem $${\displaystyle \oint \limits _{C}} g\left ( z\right ) dz=2\pi i\left ( \text{sum of residues of }g\left ( z\right ) \text{ inside }C\right )$$ But $g\left ( z\right ) =\frac{1}{\left ( z+1\right ) \left ( z+2\right ) }$ has only one pole inside $C$, which is at $z=-1$. Therefore the above becomes $$\begin{equation}{\displaystyle \oint \limits _{C}} \frac{1}{\left ( z+1\right ) \left ( z+2\right ) }=2\pi i\left ( \text{residue of }g\left ( z\right ) \text{ at }-1\right ) \tag{2} \end{equation}$$ To find residue at $-1$, we can use one of the short cuts to do that. Where we write $\frac{1}{\left ( z+1\right ) \left ( z+2\right ) }=\frac{\Phi \left ( z\right ) }{z+1}$ where $\Phi \left ( z\right )$ is analytic at $z=-1$ and $\Phi \left ( -1\right ) \neq 0$. Therefore we see that $\Phi \left ( z\right ) =\frac{1}{z+2}$. Hence residue of $\frac{1}{\left ( z+1\right ) \left ( z+2\right ) }=\Phi \left ( z_{0}\right ) =\frac{1}{\left ( -1\right ) +2}=1$.  Equation (2) becomes $${\displaystyle \oint \limits _{C}} \frac{1}{\left ( z+1\right ) \left ( z+2\right ) }=2\pi i$$ Which is same result obtained in (1) by using Cauchy integral formula directly.

To answer last part, when $R>1$, then now both poles $z=-1$ and $=-2$, are inside $C$. Therefore, we can’t split $\frac{1}{\left ( z+1\right ) \left ( z+2\right ) }$ into one part that is analytic (the $f\left ( z\right )$ in the above), in order to obtain expression $\frac{f\left ( z\right ) }{z-z_{0}}$ in order to apply Cauchy integral formula directly. Therefore when $R>1$ we should use

$${\displaystyle \oint \limits _{C}} g\left ( z\right ) dz=2\pi i\left ( \text{sum of residues of }g\left ( z\right ) \text{ inside }C\right )$$

3.2.5 Problem 5

Evaluate the integral

$${\displaystyle \oint \limits _{C}} e^{z^{2}}\left ( \frac{1}{z^{2}}-\frac{1}{z^{3}}\right ) dz$$ Where he contour is the unit circle around origin (counter clockwise direction).

Solution

$$\begin{align*}{\displaystyle \oint \limits _{C}} e^{z^{2}}\left ( \frac{1}{z^{2}}-\frac{1}{z^{3}}\right ) dz & ={\displaystyle \oint \limits _{C}} e^{z^{2}}\left ( \frac{z-1}{z^{3}}\right ) dz\\ & ={\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{3}}dz \end{align*}$$

Where $z_{0}=0$ and where

$$f\left ( z\right ) =e^{z^{2}}\left ( z-1\right )$$ But $f\left ( z\right )$ is analytic on $C$ and inside, since $e^{z^{2}}$ is analytic everywhere and $z-1$ has no poles. Hence we can use Cauchy integral formula for pole of higher order given by $${\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{\left ( z-z_{0}\right ) ^{n+1}}dz=\frac{2\pi i}{n!}f^{\left ( n\right ) }\left ( z_{0}\right )$$ Where $n=2$ in this case. Therefore, since $z_{0}=0$ the above reduces to $$\begin{equation}{\displaystyle \oint \limits _{C}} \frac{f\left ( z\right ) }{z^{3}}dz=\frac{2\pi i}{2}f^{\prime \prime }\left ( 0\right ) \tag{1} \end{equation}$$ Now we just need to find $f^{\prime \prime }\left ( z\right )$ and evaluate the result at $z_{0}=0$ $$\begin{align*} f^{\prime }\left ( z\right ) & =2ze^{z^{2}}\left ( z-1\right ) +e^{z^{2}}\\ f^{\prime \prime }\left ( z\right ) & =2e^{z^{2}}\left ( z-1\right ) +2z\left ( 2ze^{z^{2}}\left ( z-1\right ) +e^{z^{2}}\right ) +2ze^{z^{2}} \end{align*}$$

Hence

$$f^{\prime \prime }\left ( 0\right ) =-2$$ Therefore (1) becomes $$\begin{equation}{\displaystyle \oint \limits _{C}} \frac{e^{z^{2}}\left ( z-1\right ) }{z^{3}}dz=-2\pi i \tag{2} \end{equation}$$ To verify, we will do the same integration by converting it to line integration using parameterization on $\theta$.  Let $z\left ( \theta \right ) =re^{i\theta }$, but $r=1$, therefore $z\left ( \theta \right ) =e^{i\theta },dz=ie^{i\theta }d\theta$. Therefore the integral becomes $$\begin{align*}{\displaystyle \oint \limits _{C}} e^{z^{2}}\left ( \frac{z-1}{z^{3}}\right ) dz & =\int _{0}^{2\pi }e^{e^{2i\theta }}\left ( \frac{e^{i\theta }-1}{e^{3i\theta }}\right ) ie^{i\theta }d\theta \\ & =i\int _{0}^{2\pi }e^{e^{2i\theta }}\left ( \frac{e^{i\theta }-1}{e^{2i\theta }}\right ) d\theta \end{align*}$$

This is a hard integral to solve by hand. Using computer algebra software, it also gave $-2\pi i$. This verified the result. Clearly using the Cauchy integral formula to solve this problem was much simpler that using parameterization.

3.2.6 key solution to HW 2

key.pdf