How to integrate this integral using Cauchy?

Question

How can i find Solution use Cauchy Integrate?

\begin{align*} \int_{|z-1|=1}\frac{1}{z^3-1}dz \end{align*}

Answer 1

Define $f(z) = \frac{1}{z^2+z+1}$. By Cauchy integral formula we have

$\displaystyle f(1) = \frac{1}{2\pi i}\int_{|z-1|=1}\frac{f(z)}{z-1}dz = \frac{1}{2\pi i}\int_{|z-1|=1}\frac{1}{z^3-1}dz$.

Note that the other third roots of unity have negative real part, hence $f$ doesn't have a pole in $B(1,1)$.

Answer 2

So, your contour is a unit disk centered at $1$ on the complex plane.

The roots of your equation are most easily seen to be $e^{i2\pi/3}, e^{-i2\pi/3}$ and of course $1$.

Examine the roots and see if they are contained within your contour. Clearly it will contain the root $1$ and the other roots both have negative real parts.

Let $Res(f(z),z_0) = \lim_{z\rightarrow z_0} \frac{1}{3z^2}$

By residue theorem, this is $$2\pi i \frac{1}{3}$$

Which comes out to be $\frac{2\pi i}{3}$