Let C be the circle $|z|=1$. Evaluate $\int_{c}\frac{e^{2\pi z}}{(2z+1)^3}dz.$

Question

Let C be the circle $|z|=1$. Evaluate $$\int_{c}\frac{e^{2\pi z}}{(2z+1)^3}dz.$$

Any idea, suggestion, advice or solution.

Answer 1

Use Cauchy's integral theorem

$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint_C\frac{f(z)}{(z-a)^{n+1}}dz$$

with the appropiate assumptions and choices.

By the way, the result is $\;\cfrac{\pi^3i}{2e^{\pi}}\;$

Answer 2

apply $$I=i 2 \pi [Res_{z=\frac{-1}{2}}f(z)]$$

Answer 3

Use the following theorem:

https://en.wikipedia.org/wiki/Residue_theorem

I am assuming that $C$ is oriented counterclockwise. Clearly: $$f(z)=\frac{e^{2\pi z}}{(2z+1)^3}$$ is holomorphic on: $$\{z\in \mathbb{C}\mid |z|<1\}\setminus \{-\frac{1}{2}\}$$ Using the above theorem we obtain: $$\int_C\frac{e^{2\pi z}}{(2z+1)^3}dz=2\pi i\, \mathrm{res}_{z_0=-\frac{1}{2}}(\frac{e^{2\pi z}}{(2z+1)^3})$$ Now we compute: $$\mathrm{res}_{z_0=\frac{1}{2}}(\frac{e^{2\pi z}}{(2z+1)^3})=\frac{1}{2!}\lim_{z\rightarrow -\frac{1}{2}}[(\frac{e^{2\pi z}}{(2z+1)^3})(z+\frac{1}{2})^3]^{(2)}=\frac{1}{2!\cdot 8}\lim_{z\rightarrow -\frac{1}{2}}(e^{2\pi z})^{(2)}=\frac{1}{2!\cdot 8}\lim_{z\rightarrow -\frac{1}{2}}(2\pi)^2e^{2\pi z}=\frac{1}{2!\cdot 8}(2\pi)^2e^{-\pi}=\frac{\pi^2}{4e^{\pi}}$$