# Use Cauchy's integral theorem for derivatives to evaluate the contour integral.

• Aug 30th 2010, 02:34 PM
dude113
Use Cauchy's integral theorem for derivatives to evaluate the contour integral.
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.
• Aug 31st 2010, 12:44 AM
Opalg
Quote:

Originally Posted by 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.

The answer for (a) is correct. For (b), use the hint. You know that $e^z = 1+z+\frac{z^2}{2!} + \frac{z^3}{3!} + \ldots$, and so $\displaystyle\frac{e^z}{z^3} = \frac1{z^3} + \frac1{z^2} + \frac1{2!z} + \frac1{3!} + \ldots$. Now replace z by z–2 to get $\displaystyle e^{-2}\frac{e^z}{(z-2)^3} = \frac{e^{z-2}}{(z-2)^3} = \frac1{(z-2)^3} + \frac1{(z-2)^2} + \frac1{2!(z-2)} + \frac1{3!} + \ldots$.
• Aug 31st 2010, 07:38 AM
dude113
Thanks dude. (Happy)