Need some help with this. I know I can replace z4 with exp(4ix)

Let z = a + i b

[eqn] \frac{1}{1 - (a + ib)^4} = \frac{1}{1 - (a + ib)^4} \frac{1 - (a - ib)^4}{1 - (a - ib)^4} = \ldots [/eqn]
You should be able to do the rest.

[eqn] z^4 = e^{i4x} [/eqn]


[eqn] \frac{1}{1 - z^4} = \frac{1}{1 - e^{i4x}} \frac{1 - e^{-i4x}}{1 - e^{-i4x}} = \frac{1 - \cos(4x) + i \sin(4x)}{2 - 2 \cos(4x)} [/eqn]
[eqn] \Re \left( \frac{1}{1 - z^4} \right) = \frac{1}{2} [/eqn]

>>7783887

Since [math]|z|=1, {\bar z}=1/z.[/math] Let [math] w=1/(1-z^4).[/math] Then [math] Re(w)=(w+{\bar w})/2=(1/(1-z^4)+1/(1-1/z^4))/2 = 1/2. [/math]

>>7783887
funny this actually works with any p instead of 4 (if z^p is not 1)