Complex integral $\oint\frac{\sin(\pi i z/2)}{z^2-(i+1)z+i}dz$ on the circle with center $i/2$ and radius $1$

Question

Compute the complex integral $\oint\frac{\sin(\pi i z/2)}{z^2-(i+1)z+i}dz$ on the circle with center $i/2$ and radius $1$.

The denominator can be written as: $$(z-1)(z-i)$$ I want to use the CIF. So I know that $sin$ is holomorphic, so one has in relation to the CIF: $$\int_C\frac{f(z)}{(z-1)(z-i)}dz$$ where $C$ denotes the circle with center $i/2$ and radius $1$. Then I want to get the partial fractions of this. But there I didn't get the right result: $$1=A(z-i)+B(z-1) \Leftrightarrow 1= z(A+B)-Ai-B$$ So $A+B=0 \Leftrightarrow A=-B$. $$1 = Bi - B \Leftrightarrow B=\frac{1}{i-1} \Rightarrow A=\frac{-1}{i-1}$$

WolframAlpha gets for this partial fractions: $A=\frac{1}{2}+\frac{i}{2}, B=-\frac{1}{2}+\frac{i}{2}$.

If I get the right results, I have to put these into the integral and get: $$\int_C\frac{\frac{f(z)}{i-1}}{z-1}dz-\int_C\frac{\frac{f(z)}{i-1}}{z-i}dz=\frac{1}{i-1}\left(\frac{1}{2\pi i} \sin\left(-\frac{1}{2}\pi i\right)-0\right)$$

Is this correct? If yes, how can simplify the sinus in the solution?

Answer 1

Assuming that thing there under the integral sign is the circle of radius $\;1\;$ around $\;i/2\;$ , and putting

$$f(z)=\frac{\sin\frac{\pi iz}2}{z-1}\implies\oint_{|z-i/2|=1}\frac{\frac{\sin\frac{\pi iz}2}{z-1}}{z-i}dz=2\pi if(i)=2\pi i\frac{\sin\left(-\frac\pi2\right)}{i-1}=\frac{2\pi i}{1-i}=-\pi+\pi i$$