Contour integral of $\frac{\sin z}{(z^2+1)^2}$.

Question

I have the next contour integral in complex analysis: $$\oint_{\gamma} \frac{\sin (z)}{(z^2+1)^2} dz$$ With $\gamma:z=2e^{it}+1, 0\leq t \leq 2\pi$. I have tried to use the Cauchy integral formula, but I don't know how to apply the partial fractions to get factors as $\dfrac{1}{z-a}$.

Answer 1

Use the formula$$f^{(n)}(a)=\frac {n!}{2\pi i}\oint\limits_{\mathrm C}\mathrm dz\,\frac {f(z)}{(z-a)^{n+1}}$$And split the function into partial fractions. Namely$$f(z)=\frac {i\sin z}{4(z+i)}-\frac {i\sin z}{4(z-i)}-\frac {\sin z}{4(z+i)^2}-\frac {\sin z}{4(z-i)^2}$$Take the contour integral of both sides\begin{multline}\oint\limits_{\mathrm C}\mathrm dz\, f(z)=\frac i4\oint\limits_{\mathrm C}\mathrm dz\,\frac {\sin z}{z+i}-\frac i4\oint\limits_{\mathrm C}\mathrm dz\,\frac {\sin z}{z-i}-\frac 14\oint\limits_{\mathrm C}\mathrm dz\,\frac {\sin z}{(z+i)^2}\\-\frac 14\oint\limits_{\mathrm C}\mathrm dz\,\frac {\sin z}{(z-i)^2}\end{multline}Can you do the rest?

Answer 2

Both $i$ and $-i$ will be inside your contour.

$\oint \frac {f(z)}{(z-a)^2} \ dz = 2\pi i f'(a)$

$2\pi i(\frac {d}{dz} \frac {\sin z}{(z+i)^2}|_{z=i} + \frac {d}{dz} \frac {\sin z}{(z-i)^2}|_{z=-i})$

Answer 3

Hint: Use this partial fractions $$ \dfrac{1}{(z^2+1)^2}=\dfrac{1}{(z-i)^2(z+i)^2}=\dfrac14\left(\dfrac{i}{z+i}-\dfrac{i}{z-i}-\dfrac{1}{(z+i)^2}-\dfrac{1}{(z-i)^2}\right) $$