Complex Integral: Solving the Equation $\oint _{|z+i|=1} \frac{e^z}{1+z^2} dz$

[Wiemster]

Homework Statement

$$\oint _{|z+i|=1} \frac{e^z}{1+z^2} dz =?$$

The Attempt at a Solution

I substituted z+i=z' and [itex]z'=e^{i\theta}[/tex] to arrive at<br /> <br /> $$e^{-i} \int _0 ^{2 \pi} \frac{e^{e^{i \theta}}}{-ie^{i \theta}-2} d \theta$$<br /> <br /> I have no clue how to solve such an integral, any ideas??<br /> <br /> (I also did a similar exercise to arrive at the same integral but now [itex]sin(\pi/4 + exp(i \theta))[/tex] in the numerator. Are these kind of integrals analytically solvable??)[/itex][/itex]

 

[TD]

Factor the denominator: z²+1 = (z+i)(z-i), then partial fractions.
Do you know Cauchy's integral formula? It states for a inside C:

$$f(a) = {1 \over 2\pi i} \oint_C {f(z) \over z-a}\, dz$$

 

[Wiemster]

Thanks a lot! Should have thought of that of course, but now I know I can also make the others, great help!

 

[TD]

No problem

 

[Wiemster]

Well, maybe I can bother you with one more question? Most of em I can do, but there is this this one with a denominator 1+z^4 which I don't know how to separate. I tried (z^2+i)(z^2-i) but then I can't separate these...

Do you maybe have an idea?

 

[TD]

You need to find +/- sqrt(i) and +/- sqrt(-i). It factors like this:

$$\left( {z + \frac{{\sqrt 2 + \sqrt 2 i}}{2}} \right)\left( {z + \frac{{\sqrt 2 - \sqrt 2 i}}{2}} \right)\left( {z - \frac{{\sqrt 2 + \sqrt 2 i}}{2}} \right)\left( {z - \frac{{\sqrt 2 - \sqrt 2 i}}{2}} \right)$$

 

[Wiemster]

Worked like a charm! Thanks a lot!

 

[TD]

You're welcome