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Evaluate $\int\frac{1}{z^8-1}dz$ where $C := |z-1|=1$ in C.

I just don't know where to start. I have evaluated other problems like this one, but I think it's the $z^8-1$ that is throwing me off.

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  • The poles of your function are at the 8th roots of unity
  • Your curve will enclose some of these poles
  • Use the residue theorem:

$$ \int_C f(z)dz=2\pi i\sum_i\text{res}[f,z_i] $$

where $z_i$ are the poles enclosed by $C$.

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Hint: $$\int_{C}\frac{dz}{z^8-1}=\int_{C}\frac{\frac{1}{(z^4+1)(z^2+1)(z+1)}}{z-1}dz=...$$ (use Cauchy's Formula or the Residue Theorem, whichever you see fit.)

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I'm not sure that's did you get $z^8-1=(z-1)(z^4+1)$...? – icurays1 Dec 12 '12 at 20:06
@icurays1 Yeah, something is wrong with me today – Nameless Dec 12 '12 at 20:07

For this you need to apply Cauchy's Residue Theorem.

Your function, $f(z) = \frac{1}{z^8-1}$, has eight simple poles and they are given by $z^8-1=0$. These poles all lie on the unit circle and are sepreated by $\frac{2\pi}{8} = \frac{\pi}{4}$ radians. Moreover, $z=1$ is obviously a solution to $z^8-1=0$ and so the poles are the points on the unit circle, starting with $z=1$, got by rotating $\frac{\pi}{4}$ radians: $1,$ $e^{\pi i/4},$ $e^{\pi i/2},$ $e^{3\pi i/4},$ $e^{\pi i},$ $e^{5\pi i/4},$ $e^{3\pi i/2},$ or $e^{7\pi i/4}$.

Your contour $|z-1|=1$ is a circle, centred at $z=1$, with radius $1$. You need to ask yourself: How many of my poles lie on or inside my contour. If any lie on it then you're in trouble. If some lie inside it you're okay. If none lie on or inside it then the integral is exactly zero.

The unit circle - where all the poles are located - and your contour meet at the points $z=e^{\pm \pi i /3}$. To see this, draw a diagram. You get two equilateral triangles with the centres of the circles as the ends of the base points and the intersection points as the other vertices. $180^{\circ} \div 3 = 60^{\circ} \equiv \frac{\pi}{3}$ radians.

Hopefully, you can see that the poles $z = 1$, $z=e^{\pi i /4}$ and $z=e^{7\pi i /4}$ lie inside the contour, and none lie on it. By Cauchy's Residue Theorem, we know that:

$$\oint_C \frac{dz}{z^8-1} = 2\pi i\left( \text{Res}\{f,1\} + \text{Res}\{f,e^{-\pi i /4}\}+ \text{Res}\{f,e^{\pi i /4}\}\right) \, . $$

To calculate the residues, you need to use the formula for a simple pole:

$$\text{Res}\{f,z_0\} = \lim_{z\to z_0} (z-z_0)f(z) \, . $$

At this point, it might be easier to ditch polar coordinates and convert to Cartesian coordinates. The roots $z = e^{\pm \pi i /4}$ are given by

$$z = \frac{1}{\sqrt{2}} \pm i \frac{1}{\sqrt{2}} \, ,$$

moreover we can factorise $z^8-1 = (z^4-1)(z^4+1).$ Using this factorisation, the Cartesian expressions for the poles and the formula for the residues you should, after a lot of algebraic manipulation get:

$$\text{Res}\{f,1\} = \frac{1}{8} \, , $$

$$\text{Res}\{f,e^{\pm \pi i /4}\} = \frac{\sqrt{2}(1\pm i)}{16} \, . $$

Putting all of this together, we get:

$$\oint_C\frac{dz}{z^8-1} = 2\pi i \left( \frac{1}{8} + \frac{\sqrt{2}(1 - i)}{16} + \frac{\sqrt{2}(1 + i)}{16} \right) , $$

$$\oint_C\frac{dz}{z^8-1} = 2\pi i \left( \frac{1+\sqrt{2}}{8} \right) , $$

$$\oint_C\frac{dz}{z^8-1} = \pi i \left( \frac{1+\sqrt{2}}{4} \right) . $$

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