# Contour Integration

• May 4th 2011, 11:53 AM
Contour Integration
$f(z) = \frac{z+2}{2 }$ evaulate $\int c f(z) dz$ when $C$is the semicircle $z=2{e}^{ i\theta } , 0 \leqslant \theta \leqslant \pi$

I am not getting what my lecturer gets, ill show you what ive done.
• May 4th 2011, 12:02 PM
Quote:

$f(z) = \frac{z+2}{2 }$ evaulate $\int c f(z) dz$ when $C$is the semicircle $z=2{e}^{ i\theta } , 0 \leqslant \theta \leqslant \pi$

I am not getting what my lecturer gets, ill show you what ive done.

My Attempt

$dz = 2i{e}^{ i\theta } d\theta$

and then putting this in and integrating it betwen pi and 0, is this right?

If so what are other people getting for their final answer?
• May 4th 2011, 12:28 PM
Ackbeet
I get -4. What do you get?
• May 5th 2011, 10:50 AM
Quote:

Originally Posted by Ackbeet
I get -4. What do you get?

Not the same (Crying)

$\int \frac{{2e}^{ i\theta } + 2}{2 } 2i{e}^{ i\theta } d\theta = \frac{4i{e}^{ 2i\theta } + 4i{e}^{ i\theta } }{2 } d\theta = \int 2i{e}^{ 2i\theta } +2i{e}^{ i\theta }d\theta$

Is that right?
• May 5th 2011, 11:35 AM
Ackbeet
$=2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\,d \theta.$

That is what I get, which agrees with your work so far. What next?
• May 5th 2011, 11:42 AM
$= {e}^{ 2i\theta } + {2e}^{ 2i\theta }$ between $\pi$ and $0$ so ${e}^{ 2i\pi } + {2e}^{ 2i\pi} - 3$
• May 5th 2011, 11:54 AM
Ackbeet

$2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\, d\theta=2i\left(\frac{e^{2i\theta}}{2i}+\frac{e^{i \theta}}{i}\right)\Bigg|_{0}^{\pi}.$
• May 6th 2011, 03:07 AM
Quote:

Originally Posted by Ackbeet

$2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\, d\theta=2i\left(\frac{e^{2i\theta}}{2i}+\frac{e^{i \theta}}{i}\right)\Bigg|_{0}^{\pi}.$

How does that make it -4 putting the limits in?
• May 6th 2011, 04:37 AM
Ackbeet
Well, what do you get?
• May 6th 2011, 04:51 AM
Quote:

Originally Posted by Ackbeet
Well, what do you get?

${e}^{2i\pi } + 2{e}^{i\pi } -3$
• May 6th 2011, 04:54 AM
Ackbeet
Quote:

${e}^{2i\pi } + 2{e}^{i\pi } -3$

$=1-2-3\dots$
• May 6th 2011, 07:44 AM
Quote:

Originally Posted by Ackbeet
$=1-2-3\dots$

Thanks but how?
• May 6th 2011, 07:57 AM
Ackbeet
Well,

$e^{i\pi}+1=0,$ one of the most beautiful equations in all of mathematics, and also

$e^{2i\pi}=1.$

Both are points on the unit circle in the complex plane. Another way of thinking about this is that

$e^{i\theta}=\cos(\theta)+i\sin(\theta).$ Hence,

$e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+i0=-1,$ and similarly for $e^{2i\pi}.$

Make sense?
• May 6th 2011, 09:33 AM
Quote:

Originally Posted by Ackbeet
Well,

$e^{i\pi}+1=0,$ one of the most beautiful equations in all of mathematics, and also

$e^{2i\pi}=1.$

Both are points on the unit circle in the complex plane. Another way of thinking about this is that

$e^{i\theta}=\cos(\theta)+i\sin(\theta).$ Hence,

$e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+i0=-1,$ and similarly for $e^{2i\pi}.$

Make sense?

Yep, thanks.
• May 6th 2011, 09:37 AM
Ackbeet
You're welcome!