Complex Integration: Find g(2)=8πi, g(z) when |z|>3

[doubleaxel195]

Homework Statement

Let C be the circle |z|=3, described in the positive sense. Show that if

$$g(z)= \int_C \frac{2s^2-s-2}{s-z} ds$$ such that |z| does not equal 3,
then g(2)=$$8 \pi i$$. What is the value of g(z) when when |z|>3?

Homework Equations

Cauchy Integral Formula
Deformation of path

The Attempt at a Solution

I solved how to get g(2)=$$8 \pi i$$ with the Cauchy Integral Formula. But I'm not sure how to approach the second part. The only thing I can think of is that g(z) is not analytic if z=3.

 

[hunt_mat]

The answer is only non zero when all the poles are within the contour, so if they are outside...

 

[doubleaxel195]

What exactly are poles? I'm not sure we have covered that yet.

 

[hunt_mat]

poles are point at which the function is not defined, in your example the point s=z would be a pole.

 

[doubleaxel195]

Ah! I see, how silly of me. Of course it's 0 by the Cauchy-Goursat Theorem because if z is a point outside of z, g(z) becomes analytic on and within C.

 

[doubleaxel195]

Thanks.