# Cauchy's integral formula

## 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?

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## 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