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.

  • 1
    $\begingroup$ 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

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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.