complex integral - residuum theorem or something else? I have to compute integral of complex function:
$$ \int_{|z|=3} \frac{z^9}{z^{10} - 1} $$
I thought that I should after I found points where $z^{10} = 1$ use the residue theorem but I do not know how to execute that.
So I thought that maybe I should integrate by substitution $w=z^{10} $ so $ dw = z^9 dz$ which would be fortunate but then I still do not know how to compute integral over that closed curve.
Is there a simplier metod which I do not see? How to execute residuum theorem for 10 singularities of form: $\cos(k\pi/5) + i\sin(k\pi/5), k\in(0,1,2,3,4,5,6,7,8,9) $ ?
Thank you for any help!
 A: Use the argument principle: if $f(z)=z^{10}-1$; then your integral is\begin{align}\frac1{10}\int_{\lvert z\rvert=3}\frac{f'(z)}{f(z)}\,\mathrm dz&=\frac{2\pi i}{10}\times\#\{\text{zeros of $f$ in the disk }D_3(0)\}\\&=2\pi i.\end{align}
A: Residues sum to zero hence the sum of the residues from the poles on
the unit circle is minus the residue at infinity:
$$-\mathrm{Res}_{z=\infty} f(z) =
\mathrm{Res}_{z=0} \frac{1}{z^2} f\left(\frac{1}{z}\right).$$
In the present case we find with $f(z) = z^9/(z^{10}-1)$
$$\mathrm{Res}_{z=0} \frac{1}{z^2} \frac{1/z^9}{1/z^{10}-1}
= \mathrm{Res}_{z=0} \frac{1}{z^2} \frac{z}{1-z^{10}}
= \mathrm{Res}_{z=0} \frac{1}{z} \frac{1}{1-z^{10}} =  1.$$
Hence the integral is $2\pi i\times 1 = 2\pi i.$
A: There will be $10$ simple poles given by $z=e^{2πki/10} , k=0,1,2,...,9$ all lying within $\vert z\vert=3$. The residue at each pole is $\frac{z^9}{\frac{d}{dz}(z^(10)-1)}=\frac{1}{10}$.
Then $I=2πi(\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10})$ (ten times)
$\implies I=2πi$
