Cauchy's Integral Formula for a Fourth Order Singularity

[DavitosanX]

Homework Statement

This problem is from Mary L. Boas - Mathematical Methods in the Physical Sciences, Chapter 14, section 3, problem 23.

\oint_C \frac{e^{3z}dz}{(z - ln2)^{4}}

2. Homework Equations

Cauchy's integral formula

The Attempt at a Solution

First isolate the singularity:

\oint_C \frac{e^{3z}dz}{(z - ln2)^{4}} (This should read something like (e^3z/1^4)/(z-ln2)^4) but I can't seem to do it in latex...)

Then let g(z) be:

g(z) = e^{3z}

Since there's a fourth power in the singularity, g(z) must be derived three times:

g'(z) = 3e^{3z}
g''(z) = 9e^{3z}
g'''(z) = 27e^{3z}

To solve the integral, we multiply g'''(z) by:

(2 \pi i)(27e^{(3)(ln2)}) = 432 \pi i

But the book states that the answer is 72 \pi i, which is exactly 6 times less than my answer. Where did I go wrong?

 

[vela]

Homework Statement

This problem is from Mary L. Boas - Mathematical Methods in the Physical Sciences, Chapter 14, section 3, problem 23.

\oint_C \frac{e^{3z}dz}{(z - ln2)^{4}}

Homework Equations

Cauchy's integral formula

The Attempt at a Solution

First isolate the singularity:

\oint_C \frac{e^{3z}dz}{(z - ln2)^{4}} (This should read something like (e^3z/1^4)/(z-ln2)^4) but I can't seem to do it in latex...)

Then let g(z) be:

g(z) = e^{3z}

Since there's a fourth power in the singularity, g(z) must be derived three times:

You mean differentiated?

g'(z) = 3e^{3z}
g''(z) = 9e^{3z}
g'''(z) = 27e^{3z}

To solve the integral, we multiply g'''(z) by:

(2 \pi i)(27e^{(3)(ln2)}) = 432 \pi i

But the book states that the answer is 72 \pi i, which is exactly 6 times less than my answer. Where did I go wrong?

To calculate the residue, you have to divide by 1/(n-1)! where n is the order of the pole. It's part of the limit formula for calculating the residue.

 

[DavitosanX]

Thank you very much for your answer.