Solving an integral by Cauchy Formula

Question

I want to solve the integral $$\oint_{|z|=\frac{1}{2}}{\frac{e^{1-z}}{z^3(1-z)}dz}$$

Its a long time ago that I solved such integrals. Is it just by definition of the line integral? Maybe someone can help me? Thanks a lot.

Answer 1

Since the only pole of the function within the enclosed domain by the given path is zero, we get

$$\oint_{|z|=1/2}\frac{\frac{e^{1-z}}{1-z}dz}{z^3}=\left.\frac{2\pi i}{2!}\left(\frac{e^{1-z}}{1-z}\right)''\right|_{z=0}=e\pi i$$

Another way: with a few summands of the power (Laurent) series ( observe we want close to zero so $\;|z|<1\;$) :

$$\frac{e^{1-z}}{z^3(1-z)}=\frac e{z^3}\cdot e^{-z}\cdot\frac1{1-z}=\frac e{z^3}\left(1-z+\frac{z^2}2-\ldots\right)\left(1+z+z^2+\ldots\right)=$$

$$=\frac e{z^3}\left(1+\frac12z^2+\ldots\right)=\frac e{z^3}+\frac e2\frac1z+\ldots$$

and the residue is $\;\frac e2\;$ , so by the Residue Theorem

$$\oint_{|z|=1/2}\frac{e^{1-z}}{z^3(1-z)}dz=2\pi i\frac e2=e\pi i$$