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

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.

By Cauchy's integral formula, that integral is equal to $$2\pi ie^{ia}$$.
• Cauchy's integral formula says that $\int_C\frac{f(z)}{z-\omega}\,\mathrm dz=2\pi if(\omega)$, if $\omega$ belongs to the region cound by $C$. That's the case here. – José Carlos Santos Mar 19 '19 at 7:28
Let $$g(z):=e^{iz}$$. Then , by Cauchy: $$\frac{1}{2 \pi i}\int_C \frac{g(z)}{z-a} dz=g(a)=e^{ia}.$$
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.