Contour Integration

**adam_leeds** · May 4th 2011, 11:53 AM

$f(z) = \frac{z+2}{2 }$ 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.

**adam_leeds** · May 4th 2011, 12:02 PM

Originally Posted by adam_leeds

$f(z) = \frac{z+2}{2 }$ 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?

**Ackbeet** · May 4th 2011, 12:28 PM

I get -4. What do you get?

**adam_leeds** · May 5th 2011, 10:50 AM

Originally Posted by Ackbeet

I get -4. What do you get?

Not the same

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

Is that right?

**Ackbeet** · May 5th 2011, 11:35 AM

$=2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\,d \theta.$

That is what I get, which agrees with your work so far. What next?

**adam_leeds** · May 5th 2011, 11:42 AM

$= {e}^{ 2i\theta } + {2e}^{ 2i\theta }$ between $\pi$ and $0$ so ${e}^{ 2i\pi } + {2e}^{ 2i\pi} - 3$

**Ackbeet** · May 5th 2011, 11:54 AM

Your antiderivative is incorrect:

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

**adam_leeds** · May 6th 2011, 03:07 AM

Originally Posted by Ackbeet

Your antiderivative is incorrect:

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

How does that make it -4 putting the limits in?

**Ackbeet** · May 6th 2011, 04:37 AM

Well, what do you get?

**adam_leeds** · May 6th 2011, 04:51 AM

Originally Posted by Ackbeet

Well, what do you get?

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

**Ackbeet** · May 6th 2011, 04:54 AM

Originally Posted by adam_leeds

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

$=1-2-3\dots$

**adam_leeds** · May 6th 2011, 07:44 AM

Originally Posted by Ackbeet

$=1-2-3\dots$

Thanks but how?

**Ackbeet** · May 6th 2011, 07:57 AM

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?

**adam_leeds** · May 6th 2011, 09:33 AM

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.

**Ackbeet** · May 6th 2011, 09:37 AM

You're welcome!

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