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Let $\gamma$ be the semicircle $[-R,R]\cup\{z\in\mathbb{C}:|z|=R\ and\ Im{z}>0\}$, traced in the positive direction, and $R>1$. Evaluate $$\int_\gamma\frac{dz}{z^4+1}.$$

I note that $z^4+1=(z-i\sqrt{i})(z+i\sqrt{i})(z-\sqrt{i})(z+\sqrt{i}).$ Now the function in my integral is analytic in the simply connected domain that is the semicircle (except for at four isolated singularities), and the closed curve $\gamma$ does not intersect any of the singularities since $R>1$. I want to use the Residue Theorem and write

$$\int_\gamma \frac{dz}{z^4+1}=2\pi i\sum_{k=1}^4n(\gamma,z_k)Res(f;z_k).$$

At this point I need help, calculating the winding number $n$ and the residue for each of the four singularities.

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    $\begingroup$ Note that You should have only two poles inside your contour. $\endgroup$ – science Apr 16 '15 at 17:24
  • $\begingroup$ Could you explain what you mean? Aren't the moduli of each of the poles equal to 1? $\endgroup$ – nonremovable Apr 16 '15 at 17:26
  • $\begingroup$ You have $Im(z) > 0$. $\endgroup$ – science Apr 16 '15 at 17:27
  • $\begingroup$ Ah, of course. Okay in that case I think I can attack the winding number and residues for each. $\endgroup$ – nonremovable Apr 16 '15 at 17:28
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Your contour is an upper semi-circle, so only two of your poles will be inside this semi-circle.

On the other hand, the winding number denotes how many times your curve "turns" around your point $z_k$. There is a more formal definition, but in your case, the number is always 1.

As for the residues, your poles are simple, so use this formula: $$Res(f;z_k) = \lim_{z\to z_k}(z-z_k)f(z)$$

Use the decomposition of the denominator you made in the beginning. This way, you remove the expression that causes $0$ in the denominator and you just put the value of an appropriate root $z_k$ in what's left (it will be the limit of the expression).

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