Evaluate $\int_{|z|=1}\frac{e^z-1}{z^2}dz$

Question

How would you be able to evaluate $\int_{|z|=1}\frac{e^z-1}{z^2}dz$?

Are you meant to perform some integration by parts to get it in a suitable form for Cauchy's Integral formula? The only problem with that is I don't know how to do integration by parts in complex analysis.

Answer 1

Let $f(z)=e^z-1$. Then $$\oint_{|z|=1}\frac{e^z-1}{z^2}dz=2\pi i f'(0)=2\pi i \,e^0=2\pi i$$ by Cauchy's integral formula for derivatives.

Answer 2

You can also use the Residue theorem $$ \int_{|z|=1} \frac{e^z-1}{z^2} dz = 2 \pi i \, \textrm{Res}(\frac{e^z-1}{z^2}, z=0) $$

considering that $$ \lim_{z \to 0} z \cdot \frac{e^z-1}{z^2} = 1, $$

you can conclude that the residue of $\frac{e^z-1}{z^2}$ at $z=0$ is equal to $1$ and the the value of the integral.

Answer 3

Apparently, $ f(z)=\frac{e^z-1}{z^2} $ is meromorphic with $ z=0 $ a pole of order $ 2 $ of $ f(z) $ within the unit circle. Hence, we can use the Residue Theorem to compute the integral: \begin{align} \int_{|z|=1}\frac{e^z-1}{z^2}dz&=2\pi i Res(f(z), z=0)\\ &=2\pi i(e^z-1)'_{z=0}\\ &=2\pi i \end{align}

Answer 4

You can also use series expansion:

• $e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!}$
• $\Rightarrow \frac{e^z - 1}{z^2} = \frac{1}{z} + \underbrace{\sum_{n=\color{blue}{2}}^{\infty}\frac{z^{n-2}}{n!}}_{=:g(z)}$
• $g(z)$ is holomorphic in $\mathbb{C} \Rightarrow \int_{|z|=1}g(z)dz= 0$

It follows $$\int_{|z|=1}\frac{e^z-1}{z^2}dz = \int_{|z|=1}\frac{1}{z}dz \stackrel{z=e^{it}}{=} i \int_0^{2\pi}dt = 2\pi i$$