Toronto Math Forum

We have

\begin{equation}
\int_\Gamma \frac{zdz}{z^2-4z+5} 
\end{equation}

Let
\begin{equation}
f(z) = \frac{z}{z^2-4z+5} = \frac{z}{(z-(2-i))(z-(2+i))}
\end{equation}

Question a:
The point $2-i$ is outside of the contour $\Gamma$ and the point $2+i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{z}{z-2+i} \\
g(2+i) = \frac{2+i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2+i))}dz = 2\pi i g(z_0) = 2\pi i g(2+i) = 2\pi i \cdot \frac{2+i}{2i}= \pi(2+i)
\end{equation}

Question b:
The point $2+i$ is outside of the contour $\Gamma$ and the point $2-i$ is inside of the contour $\Gamma$. Then let
\begin{equation}
g(z) =\frac{z}{z-2-i} \\
g(2-i) = -\frac{2-i}{2i} \\
\int_\Gamma f(z)dz = \int_\Gamma \frac{g(z)}{(z-(2-i))}dz = 2\pi i g(z_0) = 2\pi i g(2-i) = 2\pi i \cdot -\frac{2-i}{2i}= -\pi(2-i)
\end{equation}

Question c:
Both points $2+i$ and $2-i$ are inside of the coutour $\Gamma$. Then we have
\begin{equation}
z_0 = 2+i \\
z_1 = 2-i \\
\left.Res(f;2+i) = \frac{z}{z-2+i} \right\vert_{z=2+i} = \frac{2+i}{2i} \\
\left.Res(f;2-i) = \frac{z}{z-2-i} \right\vert_{z=2-i} = - \frac{2-i}{2i}
\end{equation}
So the Residue Theorem gives us
\begin{equation}
\int_\Gamma f(z)dz = 2\pi i(\frac{2+i}{2i}-\frac{2-i}{2i}) = 2\pi i \cdot 1= 2\pi i
\end{equation}

For question a, we have
\begin{equation}
f(z)=(1-z)^{-1/2} \\
a_n = \frac{f^{(n)}(z_0)}{n!} =  \frac{f^{(n)}(0)}{n!}
\end{equation}
Then the $nth$ derivative of $f(z)$ can be derived as
\begin{equation}
f^\prime(z) = \frac{1}{2}(1-z)^{-3/2} \\
f''(z) = \frac{3}{4}(1-z)^{-5/2} \\
f'''(z) = \frac{15}{8} \times  (1-z)^{-7/2} \\
f''''(z) =\frac{105}{16} \times (1-z)^{-9/2}
\end{equation}
At $z=0$
\begin{equation}
f(0) = 1
f'(0) = \frac{1}{2} \\
f''(0) = \frac{3}{4} \\
f'''(0) = \frac{15}{8} \\
f''''(0) =  \frac{105}{16} \\
f^{(n)}(0) =  \frac{1 \times 3 \times \dots \times (2n-1)}{2^n} \\
a_n = \frac{1 \times 3 \times 5 \times \dots \times (2n-1)}{2^n n!}
\end{equation}
Thus we have the power series
\begin{equation}
f(z)= \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n = 1 + \frac{z}{2} + \frac{3z^2}{8}+ \dots
\end{equation}
The radius of convergence is
\begin{equation}
\frac{1}{R} = \lim_{n \to \infty} |\frac{a_{n+1}}{a_{n}}| \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{f^{(n+1)}(0)}{(n+1)!} \times \frac{n!}{f^{(n)}(0)} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2(n+1)} \times \frac{1}{1} \mid \\
\frac{1}{R} = \lim_{n \to \infty} \mid \frac{2n+1}{2n+2} \mid = 1 \\
R = 1
\end{equation}

For question b, let
\begin{equation}
F(z) = \arcsin(z) \\
F'(z)=\frac{1}{\sqrt{1-z^2}} = (1-z^2)^{-1/2}
\end{equation}
Note that
\begin{equation}
f(z^2)=(1-z^2)^{-1/2} \Rightarrow F'(z)=f(z^2) \\
F(z) = \int f(z^2)
\end{equation}
Then
\begin{equation}
f(z^2) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n!}z^2n \\
F(z) = \int f(z^2) = \int \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n!}z^{2n} dz
F(z) = (\sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n-1)}{2^n n! \cdot (2n+1)}z^{2n+1}) +C
\end{equation}
Since $F(0) = 0 \Rightarrow C=0$
\begin{equation}
F(z) = \sum_{n=0}^{\infty}\frac{1\times 3 \times 5 \dots \times (2n+1)}{2^n n! \cdot (2n+!)}z^{2n+1}
\end{equation}

We can first rewrite $\frac{z+2}{z+3}$ as
\begin{equation}
\frac{z+2}{z+3} = \frac{z+3-1}{z+3}
\end{equation}
\begin{align*}
\frac{z+3-1}{z+3} & = 1 - \frac{1}{z+3} \\
& = 1 - \frac{1}{z+1+2} \\
& = 1 - \frac{1}{2} \times \frac{1}{1+\frac{(z+1)}{2}} \\
& = 1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}}
\end{align*}
The power series expansion is
\begin{equation}
1 - \frac{1}{2} \times \frac{1}{1-\frac{-(z+1)}{2}} = 1 - \frac{1}{2} \times \sum_{n=0}^{\infty} (\frac{-(z+1)}{2})^n \\
= 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z-(-1))^n = 1 - \sum_{n=0}^{\infty} \frac{(-1)^n}{2^{n+1}}(z+1)^n
\end{equation}
The largest disc in which the series is valid:
\begin{equation}
\mid -\frac{z+1}{2}\mid < 1 \\
\mid z+1\mid < 2
\end{equation}

First we have
\begin{equation}
\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz
\end{equation}
The point $3$ is outside of the circle $\mid z \mid = 2$ and the point 0 is inside of the circle $\mid z \mid = 2$. Hence
\begin{equation}
\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz  = \int\limits_{\mid z \mid = 2}\frac{\frac{e^z}{z-3}}{z}\,dz
\end{equation}
This gives us function $f(z)$
\begin{equation}
f(z) = \frac{e^z}{z-3} \Rightarrow f(0) = -\frac{1}{3}
\end{equation}
So Cauchy's Formula gives us
\begin{equation}
\int\limits_{\mid z \mid = 2}\frac{e^z}{z(z-3)}\,dz =2\pi i\frac{e^0}{0-3}= -\frac{2\pi i}{3}
\end{equation}

See the attachment for the circle in this question.

My solution:
\begin{align*}
&\cos(z)=\frac{1}{2}(e^{iz}+e^{-iz})=3 \\
&(e^{iz}+e^{-iz}) = 6 \\
&e^{2iz}+1 =6e^{iz} \\
&e^{2iz}+1-6e^{iz} =0 \\
&e^{iz} = \frac{6\pm \sqrt{36-4}}{2}=3\pm 2\sqrt{2} \\
&iz=\log(3\pm 2\sqrt{2}) \\
&iz=\log(3\pm 2\sqrt{2})=\ln(3\pm 2\sqrt{2})+i2k\pi \qquad  k\in \mathbb{Z} \\
&z = -i\ln(3\pm 2\sqrt{2})+2k\pi \qquad k \in \mathbb{Z}
\end{align*}

Please reformat: \cos, \sin , \ln, \log etc