Evaluate the integral $\int_C f(z)dz$ where $C$ is the unit circle centered at the origin with $f(z) = e^{iz}/(z-a)$, $0<a<1$?

Question

Evaluate the integral $$\int_C f(x)dx \,$$ where C is the unit circle centered at C the origin with f(z) = $$\frac {e^{iz}}{z-a}\,$$ for 0 < a < 1.

I'm in complex and we are using the idea of Cauchy's theorem to evaluate the integrals. I get $$\int_C \frac{1}{z-a}dz+ \int_C \frac{iz}{z-a}dz- \int_C \frac{z^2}{2(z-a)}dz + ...\,$$ but I'm not sure if the answer is 2$\pi$i or if the following term contributes. I'm thrown off by the $z-a$ in the denominator.

Answer 1

By Cauchy's integral formula, that integral is equal to $2\pi ie^{ia}$.

Answer 2

Let $g(z):=e^{iz}$. Then , by Cauchy: $\frac{1}{2 \pi i}\int_C \frac{g(z)}{z-a} dz=g(a)=e^{ia}.$

Answer 3

If you want to do this using only Cauchy's Theorem (and not the Residue Thorem of Cauchy's integral formula) you can do the following: let $g(z)=\frac {e^{iz}-e^{ia}} {z-a}$ for $z \neq a$ and $ie^{ia}$ for $z=a$. Verify that $g$ is analytic so that $\int_Cg(z)dz=0$ by Cauchy's Theorem. Now you should be able to compute the integral of $\frac {e^{ia}} {z-a}$ directly from definition and add these two integrals to get the answer.