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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.

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

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  • $\begingroup$ Can you explain how you got that answer? $\endgroup$ – Amanda Lococo Mar 19 '19 at 7:25
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Mar 19 '19 at 7:28
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Let $g(z):=e^{iz}$. Then , by Cauchy: $\frac{1}{2 \pi i}\int_C \frac{g(z)}{z-a} dz=g(a)=e^{ia}.$

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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.

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