Find all possible values for (put into \(x+iy\) form)
Answer
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 \(n=0\).
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*}
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
\(u\left ( x,y\right ) =3x^{2}y-y^{3}.\) 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.
\(u\left ( x,y\right ) =\frac{y}{x^{2}+y^{2}}.\) 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.
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
Solution
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\[ \fbox{$\int _C\frac{1}{z^2}dz=1+i$}\]
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\[ \fbox{$\int _C\frac{1}{z^2}dz=1+i$}\]
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 \(z=-2\).
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 \[{\displaystyle \oint \limits _{C}} g\left ( z\right ) dz=2\pi i\left ( \text{sum of residues of }g\left ( z\right ) \text{ inside }C\right ) \]
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.