Complex Variables

w=(2) j^{1 / 4}
w=2(0+j)^{1 / 4}
w=2\left[e^{j(2 n+1) \pi / 2}\right]^{1 / 4} =2\left[e^{j(2 n+1) \pi / 8}\right]

For n=0, w=e^{j \pi / 8}
For n=2, w=2 e^{5 \pi j / 8}
For n=4, w=2 e^{9 \pi j / 8}

\begin{aligned} \Sigma u_n&=\Sigma \frac{n}{2^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{n+1}{2^{n+1}} \times \frac{2^n}{n}=\frac{1}{2}\\ \frac{1}{2}& \lt 2 \end{aligned}
\therefore By ratio test, \Sigma u_n is convergent.
(A) c=1, d=-1
\Sigma u_n=\Sigma \frac{1}{n} is divergent by P-test
(B) c=0.5, d=-10
\begin{aligned} \Sigma u_n&=\Sigma \frac{n^{-10}}{(0.5)^n}\\ &\text{Ratio test; }\\ \lim_{n \to \infty }\frac{u_{n+1}}{u_n}&=\lim_{n \to \infty }\frac{(n+1)^{-10}}{(0.5)^{n+1}} \times \frac{(0.5)^n}{n^{10}}\\ &=\frac{1}{0.5}=2\\ 2& \gt 1 \end{aligned}
\therefore By ratio test, \Sigma u_n is divergent.
(D) c=1,d=-2 \Sigma u_n=\Sigma \frac{n^{-2}}{(1)^n}=\Sigma \frac{1}{n^2}
\Sigma u_n is convergent by P-test.

\begin{aligned} Roots\;\; (z-1)(z+1)&=0\\ z&=\pm 1\\ \because \; z&=-1 \text{ is in contour} \end{aligned}
\begin{aligned} \Rightarrow \oint _c \frac{2^z}{z^2-1}&=\left [ \lim_{z \to -1} \frac{(z+1)2^z}{(z+1)(z-1)} \right ] \times 2 \pi i\\ &=\frac{2^{-1}}{(-1-1)} \times 2 \pi i\\ &=-\frac{1}{2} \pi i\\ \Rightarrow \;\; A&=1/2=0.5 \end{aligned}

\frac{1}{2 \pi j} \oint\left(z+\frac{1}{z}\right)^{2} d z \text { where } C \text { is }|z|=1
I=\frac{1}{2 \pi j} \int_{c}^{\left(z^{2}+1\right)^{2}} d z
z=0 lies inside the circle,
\begin{aligned} I &=\frac{1}{2 \pi j}\left[\frac{2 \pi j}{1 !} \frac{d}{d z}\left(z^{2}+1\right)^{2}\right]_{2=0} \\ &=\left[\frac{d}{d z}\left(z^{2}+1\right)^{2}\right]_{z=0} \\ &=\left[2\left(z^{2}+1\right) \times 2 z\right]_{z=0}=0 \end{aligned}

f(z) = \cos z is analytic every where.