problem with final part of this residue integral.

Question

evaluate $I=\int_0^{2\pi} \frac{d\theta}{k+sin\theta} $. where $k>1$.
is my approach below correct? I am stuck at the final step of this.
let $c$ be unit circle as $z=e^{i\theta}$ then $I=\int_c \frac{-iz^{-1}dz}{k+(\frac{z-z^{-1}}{2i})}=\int_c \frac{2dz}{z^2+2ikz-2}$ Now I am going to locate points at which the integrand is not analytic in unit circle, for $K>1$. I can not decide what are those points.
$z=\frac{-2ik\pm\sqrt{8-4k^2}}{2}$ but how can I decide what are the points in unit circle?

Answer 1

You made an error in simplifying the denominator. The constant term is $-1$, not $-2$. This yields

$$ z=\frac{-2\mathrm ik\pm\sqrt{4-4k^2}}2=\mathrm i(-k\pm\sqrt{k^2-1})\;. $$

Since $k\gt1$, exactly one of these two roots is in the unit circle, namely

$$ \mathrm i(-k+\sqrt{k^2-1})\;. $$