I have to compute $\int_{C[i-1, 1]} \frac{dz}{z^4+4}$ using residues.

The problem I'm having is locating the singularities. I know that $z^4 + 4$ will factor into $(z^2 + 2i)(z^2 - 2i)$ which will further factor into $(z + i\sqrt{2i})(z - i\sqrt{2i})(z + \sqrt{2i})(z - \sqrt{2i})$ but I'm not sure how to treat a rational power of $i$, in this case $i^{\frac{3}{2}}$

Also, I've only ever worked with singularities that were real.

  • $\begingroup$ I think it's best to think of it as looking for all the fourth roots of $-4$; generally I find that polar representation makes taking roots of complex numbers easier. Moreover, I don't quite understand your limits of integrations. Is that a circle, a line segment...? $\endgroup$ – Fimpellizieri Jun 3 '16 at 22:09
  • $\begingroup$ Here you have simple poles so the algebra is managable. Convert $-4$ to polar form and remember you may add any multiple of $2\pi i$ to the argument. $\endgroup$ – Marko Riedel Jun 3 '16 at 22:29

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