# Use Cauchy's Integral Formula to evaluate the following integrals.

Use Cauchy's Integral Formula to evaluate the following integral:

$$\int\limits_\Gamma \frac{1}{{(z-1)^3}{(z-2)^2}}dz$$ where $$\Gamma$$is a circumference of radius $4$ centered at $-2+i$ and traversed once in the positive(with respect to the interior of the disk) direction.

My thoughts on the problem:

I HAVE to use the Cauchy Integral Formula. I've been trying to decide the best way to change the expression in the integral. If I change it to:

$$\int\limits_\Gamma \frac{\frac{1}{{(z-2)^2}}}{{(z-1)^3}}$$

The point 2 is on the boundary of Gamma which means I can NOT use the formula. Are there any other ideas of ways I could change this integral to make it friendly enough to use the formula?

• Are you sure that $z=2$ is on the boundary..? (See here.) – Cameron Williams Apr 20 '15 at 18:49
• Just observe the point $z=2$ is outside of $\Gamma$ since $|z-z_0|=\sqrt{(2-(-2))^2+1^2}=\sqrt{17}>4$. Where $z_0=-2+i$ – Ángel Mario Gallegos Apr 20 '15 at 18:50
• Okay thank you! I was looking at a picture that I drew. This is a much better way of verifying if 2 is outside the boundary. – Kristin Apr 20 '15 at 18:53
• @Kristin No problem! Drawings can be very unreliable at times. I've definitely fallen into that trap before. – Cameron Williams Apr 20 '15 at 18:56
• I have a note that says: If a function f is analytic in a domain D and on the boundary of D then the integral is equal to the value of the function evaluated at z not, multiplied by two pi (i). Am I able to use this in my situation? – Kristin Apr 20 '15 at 19:02

Be careful, $2$ is not on the boundary of $\Gamma$. Then your approach is the correct one, letting $f(z)=1/(z-2)^2$, then by the General Cauchy Integral formula: $$\int_{\Gamma} \frac{f(z)}{(z-z_0)^3} = 2\pi i \frac{f''(z_0)}{2!} = \frac{6 \pi i}{(z_0-2)^4} = 6\pi i$$ since $f''(z)=6/(z-2)^4$ and $z_0=1$.