calculate $\int_{\vert z \vert = 1}\frac{e^z}{z(2z+1)^2}dz$.

Question

I need to compute

$$\int_{\vert z \vert = 1}\frac{e^z}{z(2z+1)^2}dz.$$

By Cauchy integral formula, this is

\begin{align} \int_{\vert z \vert = 1}\frac{e^z}{z(2z+1)^2}dz&=2\pi i \bigg(\frac{e^z}{(2z+1)^2}\bigg\vert_{z=0} + \frac{d}{dz}(\frac{e^z}{z})\bigg\vert_{z=-\frac{1}{2}} \bigg)\\ &=2\pi i (1+\frac{ze^z-e^z}{z^2})\bigg\vert_{z=-\frac{1}{2}}\\ &=2 \pi i (1-\frac{6}{\sqrt{e}})\\ &=2\pi i - \frac{12\pi i}{\sqrt{e}}. \end{align}

However, in the solutions manual it says $2\pi i - \frac{3\pi i}{\sqrt{e}}$. Where am I going wrong??

Answer 1

You're calculating the second residue wrong. The formula states that if $f$ has a second order pole at $c$, then

$$\mathop{\mathrm{Res}}_{z=c}f(z)=\lim_{z\to c}\frac{\mathrm{d}}{\mathrm{d}z}\bigl((z-c)^2f(z)\bigr).$$

In particular in your case you get that

$$\mathop{\mathrm{Res}}_{z=-\frac{1}{2}}\frac{e^z}{z(2z+1)^2}=\lim_{z\to-\frac{1}{2}}\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{e^z\left(z+\frac{1}{2}\right)^2}{z(2z+1)^2}\right)=\frac{1}{4}\frac{\mathrm{d}}{\mathrm{d}z}\left(\frac{e^z}{z}\right)\biggr|_{z=-\frac{1}{2}},$$

giving you the missing factor of $\frac{1}{4}$.

Answer 2

Your mistake happens in the incorrect application of the Cauchy-Formula at $\color{blue}{z_0 = -\frac 12}$. To do it correctly, you need to rewrite the function in the appropriate form $$\frac{e^z}{z\left(2z+1\right)^2} = {\color{blue}{\frac 14}} \frac{e^z}{z{\color{blue}{\left(z+\frac12\right)^2}}}$$