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Here is the integral I am dealing with:

$$\int_{|z|=1}\frac{z^{11}}{12z^{12}-4z^9+2z^6-4z^3+1}dz$$

I have been advised to make the change of variable $w=\frac{1}{z}$ which I believe results in:

$$\int_{|w|=1}\frac{-1}{w^{13}-4w^{10}+2z^7-4w^4+12w}dw$$

Now, Rouche's Theorem tells us the denominator (of the $w$ integral) has one root inside the unit disc (right?) so $w=0$ is the only pole we need to consider ?

Is this approach valid? Do I just need to calculate a residue now?

Are there alternative approaches that might work better, maybe not involving a change of variables?

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up vote 9 down vote accepted

Your integral is off by a factor of $-1$, since the variable change from $z$ to ${1 \over z}$ reverses the orientation of the circle. So what you need is $$\int_{|w| = 1} {1 \over w(w^{12} - 4w^9 + 2w^6 -4w^3 + 12)}\,dw$$ You could use Rouche but you don't have to: If $|w| \leq 1$ then $$|w^{12} - 4w^9 + 2w^6 -4w^3| \leq |w|^{12} + 4|w|^9 + 2|w|^6 + 4|w|^3$$ $$\leq 1 + 4 + 2 + 4 = 11$$ $$< 12$$ So the factor $w^{12} - 4w^9 + 2w^6 -4w^3 + 12$ is never zero. As the result the integrand just has the one pole at $w = 0$ and you can use the residue theorem for just that pole to get the value of the integral. The residue is ${\displaystyle{1 \over w^{12} - 4w^9 + 2w^6 -4w^3 + 12}}$ evaluated at ${\displaystyle w = 0}$, or ${\displaystyle{1 \over 12}}$, and the overall integral is ${\displaystyle 2\pi i * {1 \over 12} = {\pi i \over 6}}$.

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