Given:
- $f(z)=\frac{1}{(z^4+\frac{1}{2}z^2+\frac{1}{16})^2}$
- $K_1=\begin{Bmatrix}z | Im(z)=0 & \begin{vmatrix} re(z) \end{vmatrix}\leq R \end{Bmatrix}$
- $K_2=\begin{Bmatrix}Re^ {it}| 0 \leq t \leq \pi\end{Bmatrix}$
- We pass the figure clockwise
- R > 1
- $K = K_1 + K_2$
Proof: $\int_{K} f(z) dz=-4\pi$
Solution:
Singular points:
$f(z)=\frac{1}{(z^4+\frac{1}{2}z^2+\frac{1}{16})^2}=\frac{1}{(z-\frac{i}{2})^4(z+\frac{i}{2})^4}$
Residues:
Only the singular point $\frac{i}{2}$ lies in the closed area.
To calculate the residue in $z=\frac{i}{2}$ I have to calculate $\frac{g'''(\frac{i}{2})}{3!}$ with $g(z)=\frac{1}{(z+\frac{i}{2})^4}$
The residue $z=\frac{i}{2}$ appears 4 times, so I have to multiply the solution by 4.
The residue lies in the circle so next we multyply by $2\pi i$
I do not become a solution of $-4\pi$, what did I do wrong?