residue integral over a closed circle

compute $\displaystyle{\int_{\delta K_{0}} \frac{1}{z^{2}+2aiz-1}}$ where $\delta K_{0}$ is a circle of radius one around 0. And $a>1$.

the linear factors give: $\displaystyle{\frac{1}{z^{2}+aiz-1} = \frac{1}{(z-(-ai+\sqrt{a^{2}-1})(z-(-ai-\sqrt{a^{2}-1})}}$

The two residues calculated from this are: $\displaystyle{\frac{1}{2\sqrt{a^{2}-1}}}$ and $\displaystyle{\frac{1}{-2\sqrt{a^{2}-1}}}$

So the sum of them is 0. But that is not true! Since it is a circle of the form : $z=e^{i\phi}$ with $\phi\in [0,2\pi]$ Either both linear factors must lie within or none?

Does anybody see where I went wrong? Please do tell.

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You calculated the roots to the quadratic wrong, they are $-ai \pm \sqrt{1 - a^2} = -ai \pm i\sqrt{a^2 - 1}$. –  Zarrax Nov 13 '11 at 19:21
Zarrax, Thank You. –  VVV Nov 13 '11 at 19:27
@Zarrax: Your comment answers the question. If it were to be made into an answer, the question would no longer have to be in the unanswered queue. –  Mårten W Oct 4 '13 at 8:56