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I have the following contour integral (with C the positively oriented unit circle centered at the origin):

$$ \frac{-i}{4}\int_{C}\frac{\left(z^2+1\right)^2}{z\left(-z^4+3z^2-1\right)}dz $$

It has isolated singularities inside $C$ at $z = 0, \pm\sqrt{\frac{3-\sqrt{5}}{2}}$

$z = 0$ is a simple pole and the residue at that point is easily computed. I am unsure how to compute the residues at $z = \pm\sqrt{\frac{3-\sqrt{5}}{2}}$, without having to write the Laurent series?

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If $z_0$ is a pole of $f$ of order $m$, you can use the formula $$Res(f,z_0)=\frac{1}{(m-1)!} \lim_{z\to z_0} [(z-z_0)^m \cdot f(z)]^{(m-1)}$$

where $g^{(n)}$ means the $n$-th derivative of $g$.

Please ask if you need work on how that formula is obtained.

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    sorry, I can't see the difference actually, $f^{(n)}$ means the $n$-th derivative. I'll explain it anyway, thanks. – 2017-01-09
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    I am familiar with that formula, however I don't see a way to factor the denominator in such a way that the order is the power of $(z-z_0)$. (I might be missing something obvious) – 2017-01-09
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    @Tom The notation $f^{(n)}$ is pretty standard for the $n$th derivative of $f$. – 2017-01-09
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    @user405561 you can solve the equation $-z^4+3z^2-1$ by substituting $t=z^2$. – 2017-01-09
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    Indeed, that's how I arrived at $$z0 = \pm\srqt{\frac{3-\sqrt{5}}{2}}$$, I still don't know $m$, though. – 2017-01-09
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    Then, the equation $-z^4+3z^2-1$ has four roots, $\pm\sqrt\frac{3\pm\sqrt5}{2}$, so each one has order $1$. – 2017-01-09
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    The other two are ignored because they are outside the contour – 2017-01-09
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    Yes, but then the order of the pole has to be $1$. Your polynomial can be written (except for a constant) $$\left(z-\sqrt\frac{3+\sqrt5}{2}\right)\left(z-\sqrt\frac{3-\sqrt5}{2}\right)\left(z+\sqrt\frac{3+\sqrt5}{2}\right)\left(z+\sqrt\frac{3-\sqrt5}{2}\right)$$ – 2017-01-09
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    Oh wow, it was right in my face. (－‸ლ). – 2017-01-09