Integral of $\frac{e^z-1}{z^2(z-1)}$ when $0$ has winding number $2$, $1$ has winding number $-2$.

Question

I need to compute

$$\int_\gamma \frac{e^z-1}{z^2(z-1)}dz.$$

Where $0,1 \in \gamma$ where $\gamma$ winds $2$ times counterclock about $0$ and $2$ times clockwise about $1$. We have

\begin{align} \int_\gamma f(z) dz &=2 \pi i \bigg(2 \frac{d}{dz}(\frac{e^z-1}{z-1})\bigg\vert_{z=0} + (-2)\frac{e^z-1}{z^2}\bigg\vert_{z=1} \bigg)\\ &=2\pi i (-2-2(\frac{e-1}{1}))\\ &=-4\pi i - 4\pi i (e-1)\\ &=-4\pi i e \end{align}

Did I use the winding number correct in finding the result?

Answer 1

The winding number has been used correctly. A slightly different, rather detailed approach is taken here. We consider a small cycle $\gamma_1$ with center $0$, winding number $2$, not enclosing $1$ and a small cycle $\gamma_2$ with center $1$, winding number $-2$, not enclosing $0$.

We obtain \begin{align*} \color{blue}{\int_{\gamma}\frac{e^z-1}{z^2\left(z-1\right)}\,dz} &=\int_{\gamma_1}\frac{e^z-1}{z^2\left(z-1\right)}\,dz+\int_{\gamma_2}\frac{e^z-1}{z^2\left(z-1\right)}\,dz\tag{1}\\ &=\int_{\gamma_1}\frac{e^z-1}{z}\left(\frac{1}{z-1}-\frac{1}{z}\right)\,dz +\int_{\gamma_2}\frac{e^z-1}{z^2\left(z-1\right)}\,dz\tag{2}\\ &=-\int_{\gamma_1}\frac{e^z-1}{z}\cdot\frac{1}{z}\,dz +\int_{\gamma_2}\frac{e^z-1}{z^2}\cdot\frac{1}{z-1}\,dz\tag{3}\\ &=-2\pi i\cdot 2\cdot \mathrm{res}_{z=0}\left(\frac{e^z-1}{z}\cdot \frac{1}{z}\right)\\ &\qquad+2\pi i\cdot (-2)\cdot \mathrm{res}_{z=1}\left(\frac{e^z-1}{z^2}\cdot \frac{1}{z-1}\right)\tag{4}\\ &=-2\pi i \cdot 2\cdot 1+2\pi i(-2)(e-1)\\ &\,\,\color{blue}{=-4\pi i e} \end{align*} in accordance with OPs result.

Comment:

• In (1) we split the integral using cycles $\gamma_1$ and $\gamma_2$ to separately enclose the isolated singularities.

• In (2) we separate in $\int_{\gamma_1}$ the analytic part $\frac{e^z-1}{z}=1+\frac{z}{2}+\frac{z^2}{6}+\cdots$ when evaluated at $z=0$. We also do a partial fraction decomposition to separate the analytic term $\frac{1}{z-1}$ when evaluated at $z=0$.

• In (3) we simplify the integrand of $\int_{\gamma_1}$ noting the integration over a closed curve of an analytic function is zero. We also separate the analytic part in $\int_{\gamma_2}$ when evaluated at $z=1$.

• In (4) we apply the residue theorem.