Evaluate \left ( a\right ) \int _{0}^{1+i}e^{2z}dz and \left ( b\right ) \int _{C}\frac{z+4}{\left ( z^{2}+1\right ) \left ( z-1\right ) }dz where C is the circle \left \vert z\right \vert =2
Solution:
part a.
Checking if the function f\left ( z\right ) is analytic ( This is not actually needed for this part, since integration is not over a closed path).
Let z=x+iy, hence f\left ( z\right ) =e^{2z}=e^{2\left ( x+iy\right ) }\,, hence\begin{align*} f\left ( z\right ) & =e^{2x}\left ( \cos 2y+i\sin 2y\right ) \\ & =e^{2x}\cos 2y+ie^{2x}\sin 2y\\ & =u+iv \end{align*}
Hence u=e^{2x}\cos 2y,v=e^{2x}\sin 2y and\begin{align*} \frac{\partial u}{\partial x} & =2e^{2x}\cos 2y\\ \frac{\partial v}{\partial y} & =2e^{2x}\cos 2y \end{align*}
Hence \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and\begin{align*} \frac{\partial u}{\partial y} & =-2e^{2x}\sin 2y\\ \frac{\partial v}{\partial x} & =2e^{2x}\sin 2y \end{align*}
Hence \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}, therefore it is analytic function.
Complex integration can be done along any path. Using the direct path gives\begin{align*} \int _{0}^{1+i}e^{2z}dz & =\left [ \frac{e^{2z}}{2}\right ] _{0}^{1+i}\\ & =\frac{1}{2}\left ( e^{2+2i}-1\right ) \end{align*}
part b:
f\left ( z\right ) =\frac{z+4}{\left ( z^{2}+1\right ) \left ( z-1\right ) } This function can be verified to be analytic. Verified using CAS
The function f\left ( z\right ) is analytic but it does have singularity at z=1,-i,+i, therefore it is not an entire function (An entire function is analytic everywhere and has no singularity).
f\left ( z\right ) =\frac{z+4}{\left ( z-i\right ) (z+i)\left ( z-1\right ) } has all its poles inside C, therefore, using residual theory \int _{C}f\left ( z\right ) dz=2\pi i\sum \limits _{k}residual\left ( f\left ( z\right ) \right ) _{k} gives \begin{align*} \int _{C}f\left ( z\right ) dz & =2\pi i\left [ \left ( \frac{z+4}{\left ( z-i\right ) (z+i)}\right ) _{z=1}+\left ( \frac{z+4}{\left ( z-i\right ) \left ( z-1\right ) }\right ) _{z=-i}+\left ( \frac{z+4}{(z+i)\left ( z-1\right ) }\right ) _{z=i}\right ] \\ & =2\pi i\left ( \frac{1+4}{\left ( 1-i\right ) (1+i)}+\frac{-i+4}{\left ( -i-i\right ) \left ( -i-1\right ) }+\frac{i+4}{(i+i)\left ( i-1\right ) }\right ) \\ & =2\pi i\left ( \frac{5}{2}+\frac{-i+4}{-2+2i}+\frac{i+4}{-2-2i}\right ) \\ & =2\pi i\left ( \frac{5}{2}+\frac{-i+4}{-2+2i}+\frac{i+4}{-2-2i}\right ) \\ & =2\pi i\left ( \frac{5}{2}-\frac{5}{2}\right ) \\ & =0 \end{align*}
Solution
The integrand f\left ( z\right ) is \frac{\sin \left ( 2z\right ) }{\left ( z-(2-i)\right ) \left ( z-(2+i)\right ) }, hence it is poles at z_{1}=2-i and z_{2}=2+i
part a
In this case, both poles are outside C, which is unit circle centered at 0, hence \int _{C}\frac{\sin \left ( 2z\right ) }{\left ( z-(2-i)\right ) \left ( z-(2+i)\right ) }dz=0
part b
In this case, the circle is centered at 2i and its radius is 3. A plot shows the location so of poles
We see there is one pole inside C from the plot. But to determine if each pole is inside C or not, we check for \left \vert z_{i}-center\right \vert and see if this is less than the circle radius or not. We do this for each pole. For pole z_{1}=2-i we obtain \left \vert \left ( 2-i\right ) -\left ( 2i\right ) \right \vert =\left \vert 2+3i\right \vert =\sqrt{4+9}=\sqrt{13}=3.605 hence this pole is outside C as shown in the diagram. For pole z_{2}=2+i we obtain \left \vert \left ( 2+i\right ) -\left ( 2i\right ) \right \vert =\left \vert 2-i\right \vert =\sqrt{4+1}=\sqrt{5}=2.236 which is smaller that the radius of the circle C hence this pole is inside. Therefore, using the residual theorem we obtain\begin{align*} \int _{C}f\left ( z\right ) dz & =2\pi i\left [ \left ( \frac{\sin \left ( 2z\right ) }{\left ( z-(2-i)\right ) }\right ) _{z=\left ( 2+i\right ) }\right ] \\ & =2\pi i\left ( \frac{\sin \left ( 2\left ( 2+i\right ) \right ) }{\left ( \left ( 2+i\right ) -(2-i)\right ) }\right ) \\ & =2\pi i\frac{\sin \left ( 4+2i\right ) }{2i}\\ & =\pi \sin \left ( 4+2i\right ) \end{align*}
Part c
In this case, the circle is \left \vert z-\left ( 1-2i\right ) \right \vert =2\,\,, hence it is centered at point \left ( 1,-2i\right ) and its radius is 2.
For pole z_{1}=2-i we obtain \left \vert \left ( 2-i\right ) -\left ( 1-2i\right ) \right \vert =\left \vert 1+i\right \vert =\sqrt{2} hence this pole is inside C as shown in the diagram below. For pole z_{2}=2+i we obtain \left \vert \left ( 2+i\right ) -\left ( 1-2i\right ) \right \vert =\left \vert 1+3i\right \vert =\sqrt{1+9}=\sqrt{10}=\allowbreak 3.162\,3 which is larger than the radius of the circle C hence this pole is outside.
Therefore, using the residual theorem \begin{align*} \int _{C}f\left ( z\right ) dz & =2\pi i\left [ \left ( \frac{\sin \left ( 2z\right ) }{\left ( z-(2+i)\right ) }\right ) _{z=\left ( 2-i\right ) }\right ] \\ & =2\pi i\left ( \frac{\sin \left ( 2\left ( 2-i\right ) \right ) }{\left ( \left ( 2-i\right ) -(2+i)\right ) }\right ) \\ & =2\pi i\frac{\sin \left ( 4-2i\right ) }{-2i}\\ & =-\pi \sin \left ( 4+2i\right ) \end{align*}
Solution
Part (a)
The poles of f\left ( z\right ) =\frac{2z^{3}}{\left ( z-2\right ) ^{2}} are at z=2 (order 2). The following plot shows the location of the pole relative to \Gamma
The pole is inside \Gamma \,\ hence using residual we obtain \int _{C}f\left ( z\right ) dz=2\pi i\left [ \frac{1}{\left ( n-1\right ) !}\left ( \frac{d^{n-1}}{dz^{n-1}}2z^{3}\right ) _{z=2}\right ] Where n=2 in the above, since that is the order of the pole. Therefore\begin{align*} \int _{C}f\left ( z\right ) dz & =2\pi i\left [ \left ( \frac{d}{dz}2z^{3}\right ) _{z=2}\right ] \\ & =\left ( 2\pi i\right ) \left ( 6z^{2}\right ) _{z=2}\\ & =\left ( 2\pi i\right ) \left ( 6\left ( 2\right ) ^{2}\right ) \\ & =48\pi i \end{align*}
Part b
The poles of f\left ( z\right ) =\frac{\cos \left ( z-i\right ) }{\left ( z+2i\right ) ^{3}}. The poles are at z=-2i of order 3. Therefore \Gamma includes the pole inside it, and we can use the residual, hence \int _{C}f\left ( z\right ) dz=2\pi i\left [ \frac{1}{\left ( n-1\right ) !}\left ( \frac{d^{n-1}}{dz^{n-1}}\cos \left ( z-i\right ) \right ) _{z=-2i}\right ] Where n=3 in the above, since that is the order of the pole. Therefore\begin{align*} \int _{C}f\left ( z\right ) dz & =2\pi i\left [ \frac{1}{2}\left ( \frac{d^{2}}{dz^{2}}\cos \left ( z-i\right ) \right ) _{z=-2i}\right ] \\ & =2\pi i\left [ \frac{1}{2}\left ( \frac{d}{dz}\left ( -\sin \left ( z-i\right ) \right ) \right ) _{z=-2i}\right ] \\ & =2\pi i\left [ \frac{1}{2}\left ( -\cos \left ( z-i\right ) \right ) _{z=-2i}\right ] \\ & =2\pi i\frac{1}{2}\left ( -\cos \left ( -2i-i\right ) \right ) \\ & =-\pi i\cos \left ( -3i\right ) \end{align*}
Or \int _{C}f\left ( z\right ) dz=-i\pi \cosh \left ( 3\right )
Part c
f\left ( z\right ) =\left ( z-i\right ) ^{2} has no poles. Hence if we can show it is analytic, then the closed path integral will be zero by Cauchy theorem. We can verify also it is analytic. Let z=x+iy, hence \begin{align*} f\left ( z\right ) & =\left ( \left ( x+iy\right ) -i\right ) ^{2}\\ & =x^{2}+2ixy-2ix-y^{2}+2y-1\\ & =\left ( x^{2}-y^{2}-1\right ) +i\left ( 2xy-2x\right ) \end{align*}
\,Therefore u=x^{2}-y^{2}-1 and v=2xy-2x and\begin{align*} \frac{\partial u}{\partial x} & =2x\\ \frac{\partial v}{\partial y} & =2x \end{align*}
Hence \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y} and\begin{align*} \frac{\partial u}{\partial y} & =-2y\\ \frac{\partial v}{\partial x} & =2y-2 \end{align*}
Hence \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}, therefore it is analytic function and the integral is zero.
Expand f\left ( z\right ) =\frac{1}{z^{2}+3z+2} in Taylor series. (a) About point z=0 (b) about point z=2. Determine the radius of convergence for each case.
Solution:
Taylor series is defined as f\left ( z\right ) =f\left ( a\right ) +\left ( z-a\right ) f^{\prime }\left ( a\right ) +\frac{\left ( z-a\right ) ^{2}}{2!}f^{\prime \prime }\left ( a\right ) +\cdots +\frac{\left ( z-a\right ) ^{n-1}}{\left ( n-1\right ) !}f^{\left ( n-1\right ) }\left ( a\right ) +R_{n} However, we do not calculate the series directly from the above definition. Instead, using power series method, we can formulate this to finding the series f\left ( z\right ) =\sum \limits _{n=0}^{\infty }c_{n}\left ( z-z_{0}\right ) ^{n} Where z_{0} is the point to expand around. In this case it is zero. We need to find c_{n}.
To find radius of convergence \rho , we draw a circle, centered at the point of expansion, and extend it all the way to the first pole. f\left ( z\right ) =\frac{1}{\left ( z+2\right ) \left ( z+1\right ) } hence it has a pole at z=-2 and z=-1\,.therefore, \rho =1 Here is a diagram
Where the first geometric series converges for \left \vert z\right \vert <1 and the second converges for \left \vert \frac{z}{2}\right \vert <1 or \left \vert z\right \vert <2. The series is\begin{align*} f\left ( z\right ) & =\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}z^{n}-\frac{1}{2}\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{z}{2}\right ) ^{n}\\ & =\left ( 1-z+z^{2}-z^{3}+z^{4}+\cdots \right ) -\frac{1}{2}\left ( 1-\frac{z}{2}+\frac{z^{2}}{2^{2}}-\frac{z^{3}}{2^{3}}+\frac{z^{4}}{2^{4}}+\cdots \right ) \\ & =\left ( 1-z+z^{2}-z^{3}+z^{4}+\cdots \right ) +\left ( -\frac{1}{2}+\frac{z}{2^{2}}-\frac{z^{2}}{2^{3}}+\frac{z^{3}}{2^{4}}-\frac{z^{4}}{2^{5}}+\cdots \right ) \\ & =\frac{1}{2}-z\left ( 1-\frac{1}{2^{2}}\right ) +z^{2}\left ( 1-\frac{1}{2^{3}}\right ) -z^{3}\left ( 1-\frac{1}{2^{4}}\right ) +\cdots \\ & =\frac{1}{2}-\frac{3}{4}z+\frac{7}{8}z^{3}-\frac{15}{16}z^{3}+\cdots \end{align*}
f\left ( z\right ) =\sum \limits _{n=0}^{\infty }c_{n}\left ( z-z_{0}\right ) ^{n} Where z_{0} is the point to expand around. In this case it is 2.
To find radius of convergence \rho , we make a circle, centered at the point of expansion, and extend it all the way to the first pole. Therefore radius of convergence is \rho =3 As shown in this diagram
We need to find c_{n}. Writing \begin{align*} f\left ( z\right ) & =\frac{1}{z^{2}+3z+2}=\frac{1}{\left ( z+2\right ) \left ( z+1\right ) }\\ & =\frac{1}{z+1}-\frac{1}{z+2}\\ & =\frac{1}{\left ( z-2\right ) +3}-\frac{1}{\left ( z-2\right ) +4}\\ & =\frac{1}{3}\frac{1}{1+\left ( \frac{z-2}{3}\right ) }-\frac{1}{4}\frac{1}{1+\left ( \frac{z-2}{4}\right ) }\\ & =\frac{1}{3}\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{z-2}{3}\right ) ^{n}-\frac{1}{4}\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{z-2}{4}\right ) ^{n} \end{align*}
Where the first geometric series converges for \left \vert \frac{z-2}{3}\right \vert <1 or \left \vert z-2\right \vert <3 or \left \vert z\right \vert <5 and the second converges for \left \vert \frac{z-2}{4}\right \vert <1 or \left \vert z-2\right \vert <4 or \left \vert z\right \vert <6. And the series is\begin{align*} f\left ( z\right ) & =\frac{1}{3}\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{z-2}{3}\right ) ^{n}-\frac{1}{4}\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{z-2}{4}\right ) ^{n}\\ & \\ & =\frac{1}{3}\left ( 1-\frac{1}{3}\left ( z-2\right ) +\left ( \frac{1}{3}\right ) ^{2}\left ( z-2\right ) ^{2}-\left ( \frac{1}{3}\right ) ^{2}\left ( z-2\right ) ^{3}+\cdots \right ) \\ & -\frac{1}{4}\left ( 1-\frac{1}{4}\left ( z-2\right ) +\left ( \frac{1}{4}\right ) ^{2}\left ( z-2\right ) ^{2}-\left ( \frac{1}{4}\right ) ^{3}\left ( z-2\right ) ^{3}+\cdots \right ) \\ & \\ & =\left ( \frac{1}{3}-\frac{1}{9}\left ( z-2\right ) +\frac{1}{27}\left ( z-2\right ) ^{2}-\frac{1}{27}\left ( z-2\right ) ^{3}+\cdots \right ) \\ & -\left ( \allowbreak \frac{1}{4}-\frac{1}{16}\left ( z-2\right ) +\frac{1}{64}\left ( z-2\right ) ^{2}-\frac{1}{256}\left ( z-2\right ) ^{3}+\cdots \right ) \\ & \\ & =\frac{1}{12}-\frac{7}{144}\left ( z-2\right ) +\frac{37}{1728}\left ( z-2\right ) ^{2}-\frac{229}{6912}\left ( z-2\right ) ^{3}+\cdots \end{align*}
solution
part (a)
f\left ( z\right ) =\frac{1}{2+z} around point 1-8i.
To find radius of convergence \rho , we make a circle, centered at the point of expansion, and extend it all the way to the first pole. f\left ( z\right ) =\frac{1}{2+z} hence it has a pole at z=-2, Therefore, \rho =\sqrt{3^{2}+8^{2}}=\sqrt{73}
As seen by this diagram.
Now the Taylor series is found
f\left ( z\right ) =\sum \limits _{n=0}^{\infty }c_{n}\left ( z-z_{0}\right ) ^{n} Where z_{0} is the point to expand around. In this case it is 1-8i. We need to find c_{n}. Writing \begin{align*} f\left ( z\right ) & =\frac{1}{2+z}=\frac{1}{2+z-\left ( 1-8i\right ) +\left ( 1-8i\right ) }\\ & =\frac{1}{\left ( 3-8i\right ) +\left ( z-\left ( 1-8i\right ) \right ) }\\ & =\frac{1}{\left ( 3-8i\right ) }\frac{1}{1+\frac{1}{3-8i}\left ( z-\left ( 1-8i\right ) \right ) }\\ & =\frac{1}{\left ( 3-8i\right ) }\sum \limits _{n=0}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{1}{3-8i}\right ) ^{n}\left ( z-\left ( 1-8i\right ) \right ) ^{n} \end{align*}
Hence\begin{align*} f\left ( z\right ) & =\frac{1}{\left ( 3-8i\right ) }\left ( 1-\left ( \frac{1}{3-8i}\right ) \left ( z-\left ( 1-8i\right ) \right ) +\left ( \frac{1}{3-8i}\right ) ^{2}\left ( z-\left ( 1-8i\right ) \right ) ^{2}+\cdots \right ) \\ & =\frac{1}{\left ( 3-8i\right ) }-\frac{1}{\left ( 3-8i\right ) ^{2}}\left ( z-\left ( 1-8i\right ) \right ) +\frac{1}{\left ( 3-8i\right ) ^{3}}\left ( z-\left ( 1-8i\right ) \right ) ^{2}+\cdots \\ & =\left ( \frac{3}{73}+\frac{8i}{73}\right ) +\left ( \frac{55}{5329}-\frac{48i}{5329}\right ) \left ( z-\left ( 1-8i\right ) \right ) -\left ( \frac{549}{389017}+\frac{296i}{389017}\right ) \left ( z-\left ( 1-8i\right ) \right ) ^{2}+\cdots \end{align*}
part (b)
f\left ( z\right ) =1+\frac{1}{2+z^{2}} around point i. In this case, the center of circle is at i and hence \rho =\sqrt{1^{2}+2^{2}}=\sqrt{5} As shown in this diagram
And Taylor series is \begin{align*} f\left ( z\right ) & =1+\frac{1}{2+z^{2}}\\ & =1+\frac{1}{\left ( z-i\sqrt{2}\right ) \left ( z+i\sqrt{2}\right ) } \end{align*}
Doing partial fractions gives f\left ( z\right ) =1+\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}-iz}+\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}+iz} Convert so that z-i appears. iz=i\left ( z-i\right ) -1, hence\begin{align} f\left ( z\right ) & =1+\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}-\left ( i\left ( z-i\right ) -1\right ) }+\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}+\left ( i\left ( z-i\right ) -1\right ) }\nonumber \\ & =1+\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}-i\left ( z-i\right ) +1}+\frac{1}{2\sqrt{2}}\frac{1}{\sqrt{2}+i\left ( z-i\right ) -1}\nonumber \\ & =1+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) -i\left ( z-i\right ) }+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) +i\left ( z-i\right ) }\nonumber \\ & =1+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) }\frac{1}{1-\frac{i}{\left ( \sqrt{2}+1\right ) }\left ( z-i\right ) }+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) }\frac{1}{1+\frac{i}{\left ( \sqrt{2}-1\right ) }\left ( z-i\right ) }\tag{1} \end{align}
Now geometric series can be used.\begin{align*} \frac{1}{1-\frac{i}{\left ( \sqrt{2}+1\right ) }\left ( z-i\right ) } & =\sum _{n}^{\infty }\left ( \frac{i}{\left ( \sqrt{2}+1\right ) }\right ) ^{n}\left ( z-i\right ) ^{n}\\ & =1+\frac{i}{\sqrt{2}+1}\left ( z-i\right ) -\frac{1}{\left ( \sqrt{2}+1\right ) ^{2}}\left ( z-i\right ) ^{2}-\frac{i}{\left ( \sqrt{2}+1\right ) ^{3}}\left ( z-i\right ) ^{3}+\frac{1}{\left ( \sqrt{2}+1\right ) ^{4}}\left ( z-i\right ) ^{4}\cdots \end{align*}
And\begin{align*} \frac{1}{1+\frac{i}{\left ( \sqrt{2}-1\right ) }\left ( z-i\right ) } & =\sum _{n}^{\infty }\left ( -1\right ) ^{n}\left ( \frac{i}{\left ( \sqrt{2}-1\right ) }\right ) ^{n}\left ( z-i\right ) ^{n}\\ & =1-\frac{i}{\sqrt{2}-1}\left ( z-i\right ) +\frac{1}{\left ( \sqrt{2}-1\right ) ^{2}}\left ( z-i\right ) ^{2}-\frac{i}{\left ( \sqrt{2}-1\right ) ^{3}}\left ( z-i\right ) ^{3}+\frac{1}{\left ( \sqrt{2}-1\right ) ^{4}}\left ( z-i\right ) ^{4}\cdots \end{align*}
Substituing the above into Eq. (1) gives\begin{align*} f\left ( z\right ) & =1+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) }\left ( 1+\frac{i}{\sqrt{2}+1}\left ( z-i\right ) -\frac{1}{\left ( \sqrt{2}+1\right ) ^{2}}\left ( z-i\right ) ^{2}-\frac{i}{\left ( \sqrt{2}+1\right ) ^{3}}\left ( z-i\right ) ^{3}+\frac{1}{\left ( \sqrt{2}+1\right ) ^{4}}\left ( z-i\right ) ^{4}\cdots \right ) \\ & +\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) }\left ( 1-\frac{i}{\sqrt{2}-1}\left ( z-i\right ) +\frac{1}{\left ( \sqrt{2}-1\right ) ^{2}}\left ( z-i\right ) ^{2}+\frac{i}{\left ( \sqrt{2}-1\right ) ^{3}}\left ( z-i\right ) ^{3}-\frac{1}{\left ( \sqrt{2}-1\right ) ^{4}}\left ( z-i\right ) ^{4}\cdots \right ) \end{align*}
Looking at coefficient of term \left ( z-i\right ) ^{n} power. For n=0 power the term is \begin{align*} c_{0} & =1+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) }+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) }\\ & =2 \end{align*}
And for c_{1} it is\begin{align*} c_{1} & =\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) }\frac{i}{\sqrt{2}+1}-\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) }\frac{i}{\sqrt{2}-1}\\ & =-2i \end{align*}
And for c_{2}\begin{align*} c_{2} & =-\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) }\frac{1}{\left ( \sqrt{2}+1\right ) ^{2}}-\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) }\frac{1}{\left ( \sqrt{2}-1\right ) ^{2}}\\ & =-5 \end{align*}
And for c_{3}\begin{align*} c_{2} & =-\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}+1\right ) }\frac{i}{\left ( \sqrt{2}+1\right ) ^{3}}+\frac{1}{2\sqrt{2}}\frac{1}{\left ( \sqrt{2}-1\right ) }\frac{i}{\left ( \sqrt{2}-1\right ) ^{3}}\\ & =12i \end{align*}
and so on. Hence the series is\begin{align*} f\left ( z\right ) & =c_{0}+c_{1}\left ( z-i\right ) +c_{2}\left ( z-i\right ) ^{2}+c_{3}\left ( z-i\right ) ^{3}+\cdots \\ & =2-2i\left ( z-i\right ) -5\left ( z-i\right ) ^{2}+12i\left ( z-i\right ) ^{3}+\cdots \end{align*}