Evaluate $\int\frac{e^z-1}{z}\mathrm dz$ along the unit circle

Question

How do I evaluate the following integral? $$\oint_{C}\frac{e^z-1}{z}\mathrm dz$$ where $C$ is the unit circle (counter-clockwise).

I have just learned Cauchy-Goursat's Theorem, but I cannot apply it here since there is a singularity at $z=0$. So I try to go brute force and evaluate the contour integral as follows: $$\int_{0}^{1}\dfrac{e^{e^{it}}-1}{e^{it}}ie^{it}\,\mathrm dt = \int_{0}^{1}({e^{e^{it}}-1})i\,\mathrm dt .$$ But then I'm stuck again as to how to evaluate this.

Thanks!

Answer 1

There is really no singularity at $0$: let $g(z)=1+\frac z {2!}+\frac {z^{2}} {3!}+...$. Then Goursat theorem can be aplied to $g$. But $f$ and $g$ are one and the same functions on the unit circle so the integral is $0$.

Answer 2

Since$$\frac{e^z-1}z=\frac{z+\frac1{2!}z^2+\frac1{3!}z^3+\cdots}z=1+\frac1{2!}z+\frac1{3!}z^2+\cdots,$$the integral is $0$.