# Homework Help: Complex Contour Integral 2

1. May 1, 2012

### bugatti79

1. The problem statement, all variables and given/known data
Evaluate each of the following by Cauchy's Integral formula

a)$\int_cj \frac{\cos z}{3z-3\pi} dz$ c1: |z|=3, c2:|z|=4

b) $\int_c \frac{e^{3z}}{z-ln(2)} dz$ c=square with corners at $\pm(1\pm i)$

2. Relevant equations
$f(z_0)=\frac{1}{2 \pi i}\int_c \frac{f(z)}{z-z_0}$

3. The attempt at a solution

a) $\int_{c_j} \frac{\cos z}{3z-3\pi} dz$ c1: |z|=3, c2:|z|=4

$\frac{1}{3}\int_{c1} \frac{\cos z}{z- \pi} dz =0$ since $\pi$ lies outside c1 and hence $\frac{\cos z}{z- \pi}$ is analytic on and inside c1

$\frac{1}{3}\int_{c1} \frac{\cos z}{z- \pi} dz = \frac{1}{3} (2\pi i) \cos (\pi)= -\frac{2}{3} \pi i$ since $\pi$ lies inside c2

b) $\int_c \frac{e^{3z}}{z-ln(2)} dz$ c=square with corners at $\pm(1\pm i)$

$\int_c \frac{e^{3z}}{z-ln(2)} dz=2 \pi i e^{3 ln(2)}=2^4 \pi i$...?

Thanks

2. May 1, 2012

### jackmell

Looks ok to me.

3. May 2, 2012

### bugatti79

In b) if we had $\int_c \frac{e^{3z}}{(z-ln(2))^3} dz$

We'd get the same answer because we have $(z-ln(2))^3=0 \implies z-ln(2)=0$..?

4. May 2, 2012

### jackmell

No.
$$f^{(n)}(a)=\frac{n!}{2\pi i}\oint \frac{f(z)}{(z-a)^{n+1}}dz$$

5. May 3, 2012

### bugatti79

$\displaystyle \oint \frac{e^{3z}}{(z-ln(2))^{3}}dz=\frac{2\pi i}{3!} (3e^{3(ln 2)})=2^4 \pi i$..which is the same as original q part b)?

6. May 3, 2012

### jackmell

When I plug your integral into that formula, I get:

$$\frac{d^2}{dz^2}\left(e^{3z}\right)\biggr|_{z=\ln(2)}=\frac{2!}{2\pi i} \oint \frac{e^{3z}}{(z-\ln(2))^3}dz$$

or

$$\oint \frac{e^{3z}}{(z-\ln(2))^3}dz=72\pi i$$

Also, try and learn to check them in Mathematica:

Code (Text):

NIntegrate[(Exp[3*z]/(z - Log[2])^3)*2*I*
Exp[I*t] /. z -> 2*Exp[I*t], {t, 0, 2*Pi}]
N[72*Pi*I]

-1.4210854715202004*^-14 + 226.19467105847158*I

0. + 226.1946710584651*I

7. May 3, 2012

### bugatti79

I didnt differentiate twice!

Thank you!