Question

for anyone who knows complex analysis, or even is good with line integrals:

\[\int\limits_{|z+1|=2}^{} z^{2} / (4-z^2)\]

Answer 1

the answer is \[2i \pi \] but I don't know how they got that

Answer 2

it's based on Cauchy's Integral Formula:

\[\int\limits_{\gamma}^{} f(z)/(z-z_{0}) = 2i \pi f(z_0)\]

Answer 3

never mind, figured it out

Answer 4

\[\frac{z^2}{4-z^2}=\frac{z^2}{(2-z)(2+z)}\quad\Rightarrow\quad z_{p1}=2\quad,\quad z_{p2}=-2\,\,\,\mbox{(poles of complex function)}\]
\[\gamma\,:\,|z+1|=2\,\,\,\mbox{(circle in complex plane, center=(-1,0) , radius=2)}\]
\[z_{p1}=2\not\in\gamma\,\,,\,\,z_{p2}=-2\in\gamma\quad\Rightarrow\quad\int\limits_{\gamma}\underbrace{\frac{z^2}{2-z}}_{f(z)}\cdot\frac{1}{\underbrace{2}_{-z_0}+z}dz=2i\pi f(z_0)=\]
\[=2i\pi f(-2)=2i\pi\frac{(-2)^2}{2-(-2)}=2i\pi\]