Evaluate $\int_{|z|=1} \frac{1}{e^z -1-2z}dz$ by using Cauchy's residue theorem

Question

Evalulate the integral by using Cauchy's residue theorem $$\int_{|z|=1} \frac{1}{e^z -1-2z}dz$$

My attempt: $ f(z) =\frac{1}{e^z -1-2z}$, now put $z= 1$ we get $f(z)=-1$

so $$\int_{|z|=1} \frac{1}{e^z -1-2z}dz= -2\pi i$$

Is it correct?

Answer 1

First of all, it is clear that $0$ is a zero of $e^z-1-2z$. Is it the only zero in the closed unit circle? Yes, but that is not obvious. Suppose that $\lvert z\rvert=1$. Then\begin{align}\lvert e^z-1\rvert&=\left\lvert z+\frac{z^2}{2!}+\frac{z^3}{3!}+\cdots\right\rvert\\&\leqslant\lvert z\rvert+\frac{\lvert z\rvert^2}{2!}+\frac{\lvert z\rvert^3}{3!}+\cdots\\&=1+\frac1{2!}+\frac1{3!}+\cdots\\&=e-1\\&<2\\&=\lvert2z\rvert\end{align}So, $e^z-1-2z$ has no zeros in the unit circle. Since $2z$ has exactly one zero (counting with multiplicities) in the open unit circle, it follows from Rouché's theorem that $e^z-1-2z$ has exactly one zero in that region. And we already not that that zero is $0$. So, your function has a single (and simple) pole in that disk and therefore the residue theorem says that\begin{align}\int_{\lvert z\rvert=1}\frac1{e^z-1-2z}\,\mathrm dz&=2\pi i\operatorname{res}_{z=0}\left(\frac1{e^z-1-2z}\right)\\&=2\pi i\frac1{e^0-2}\\&=-2\pi i.\end{align}