Cauchy's residue application problem

Question

I would like to know what did I do wrong. There's my problem :

I= $\frac{1}{2\pi i}$ $\int_a \frac{1}{z^4+1}~dz$

Where a $x^2 + y^2 = 2x$

I already know: there're 4 poles, but only 2 fits for me (right ones): $e^\frac{i\pi}{4}$ and $e^\frac{-i\pi}{4}$

I've both residues: -$\dfrac{1}{4}$$e^\frac{i\pi}{4}$ and $\dfrac{1}{4}$$e^\frac{-i\pi}{4}$

$\sum\limits_{b=1}^2 z_b$ = $\dfrac{ -\sqrt{2}}{4}$i

But correct answer is given without i. What's wrong?

Thanks.

If that's so then the imaginary part of the residues cancel out as you have typed and when you multiply with $2\pi i$ you should have imaginary result. — Harto Saarinen, Nov 24 '15 at 20:10
According to my calculations the answer should be $\frac{\pi i}{\sqrt{2}}$. At least you forgot to multiply by $2 \pi i$. — Harto Saarinen, Nov 24 '15 at 20:14
@HartoSaarinen Sorry, I didn't get your mind. Correct answer is given without imaginary result. — Edgar, Nov 24 '15 at 20:15
@HartoSaarinen Can you give me extended solution ? I can't f ind my mistake :) BTW answer is negative — Edgar, Nov 24 '15 at 20:18
You have wrong poles. You took the ones from the right. You should take them inside the circle which is in left half-plane. — Harto Saarinen, Nov 24 '15 at 20:19
Sry for the confusion. You should take them from RIGHT side of the plane. And calculating the residue of those above and adding up gives you the correct result. (which indeed is negative). You can pretty much forget everything above. Messed up with left and right lol. — Harto Saarinen, Nov 24 '15 at 20:27

score 1 · Accepted Answer · answered Nov 24 '15 at 20:33

Your contour $C:$ $|z-1|=1$ is in the right half-plane. The poles of integrand are $-e^{i \pi/4}, -e^{3 i \pi/4}, e^{i \pi/4}$ and $e^{3 i \pi/4}$.

Two of those are inside $C$: $e^{i \pi/4}$ and $-e^{3 i \pi/4}$. Calculating the residues and summing them gives $-\frac{\sqrt{2}}{4}$.

1 Answers1