$\int_{C}{\sin(\bar{z})}\,dz$ where C is the path connecting $-i$ to $i$ along the unit circle in the positive direction

Question

I need help calculating $\int_{C}{\sin(\bar{z})}\,dz$ where C is the path connecting $-i$ to $i$ along the unit circle in the positive direction.

I have already noticed that for $|z|=1$, $\bar{z} = \frac{1}{z}$ so that $\int_{C}\sin({\bar{z})}\,dz = \int_{C}\sin\frac{1}{z}\,dz$

Also, $\operatorname{Res}_{z=0}\sin(\frac{1}{z})=1$ so that $\int_{|z|=1}\sin\frac{1}{z}\,dz = 2\pi i$ by the Cauchy Residue Theorem.

I just can't figure out how to relate the two. Any help is appreciated, Thanks!

Answer 1

I am no expert, just a student taking some complex analysis course, but in order to use the residue formula you need to integrate over a closed loop. So you should start by closing the loop via another path and then the integral over C and the ¨closing path¨ integral should add up to the value you calculated with the residue formula.

You could probably do it that way, but the easiest approach to me would be to directly apply the parametrization of the path C onto the integral, which can be solved pretty easily and yields a cosine function.