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**dude113** a) **Use Cauchy's integral theorem for derivatives to evaluate the contour integral**

I = e^z/((z-2)^3) dz around the contour |z|=3.

b) **Verify that your answer is correct by finding a suitable Laurent series expansion of e^z/((z-2)^3) and integrating this series around the contour |z|=3.**

[**Hint**: e^z= e^2.e^z-2]

a) I've got k=2, f(z)=e^z and z_0=2.

f'(z) = e^z so f'(z_0)= e^2. Equally f''(z_0) = e^2. (Right?)

Then I = (2*pi*i/2!)*f''(z_0) = (pi*i)*e^2

= pi*i*e^2 as my final answer.

For part b) I've wrote out the expansion of e_z first. Then divided through by ((z-2)^3). But I'm struggling on where to go next.

The reason I'm so rusty is this paper is a rework and so its been a few months since I've done any calculus work. If anyone could give me a hand, I'd appreciate it.