# Math Help - Contour Integration

1. ## 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.

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

3. I get -4. What do you get?

4. 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?

5. $=2i\int_{0}^{\pi}(e^{2i\theta}+e^{i\theta})\,d \theta.$

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

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

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

8. 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?

9. Well, what do you get?

10. Originally Posted by Ackbeet
Well, what do you get?
${e}^{2i\pi } + 2{e}^{i\pi } -3$

${e}^{2i\pi } + 2{e}^{i\pi } -3$
$=1-2-3\dots$

12. Originally Posted by Ackbeet
$=1-2-3\dots$
Thanks but how?

13. 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?

14. 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.

15. You're welcome!