# Poles of a rationalfunction

Consider the function:

$f(z) = \dfrac{1}{z^{4}+z^{2}+1}$

I want to find the poles. I have deduced that there are 4 simple poles and 2 of them are in the upper half-plane. I'm not sure how I can find them, I have tried to find the zeros of $z^{4}+z^{2}+1$, but with no luck. I have also tried to see if I can factorize $z^{4}+z^{2}+1$, but I'm still stuck.

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First find the zeros of $z^2 + z + 1$.. – Cocopuffs Nov 4 '12 at 9:35

You have a quadratic in $z^2$. Write $x=z^2$ to get $$x^2 + x + 1$$ from the quadratic formula, you get the roots as $$x = \frac{-1 \pm i\sqrt{3}}{2}$$ Now find the square roots for each to recover the four roots of $z$.

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A way to find the factorization is:

$z^4+z^2+1=(z^4+2z^2+1)-z^2 = (z^2+1)^2-(z)^2= (z^2+1-z)(z^2+1+z)$

Then, you can just apply the quadratic formula for each factor.

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\begin{align} z^4+z^2+1 &= (z^2-z+1)(z^2+z+1) \\ &= ((z-1/2)^2+3/4)((z+1/2)^2+3/4) \\ &= (z-1/2+\sqrt{3}i/2)(z-1/2-\sqrt{3}i/2)(z+1/2+\sqrt{3}i/2)(z+1/2-\sqrt{3}i/2) \\ \end{align}

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Let's find the zeroes of $p(z) = z^4 + z^2 + 1$. Letting $\zeta = z^2$, we must first solve $\zeta^2 + \zeta + 1 = 0$, which gives $\zeta_{1,2} = -\frac 12 \pm \sqrt{\frac 14 - 1} = \frac{-1 \pm \sqrt 3i}2$ so the four roots of $p$ are the square roots of $\zeta_{1,2}$.

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