Need some help with this. I know I can replace...

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Need some help with this. I know I can replace z4 with exp(4ix) which gives cos(4x)+isin(4x).

The most I am able to figure out is maybe replace z4 with the cos/sin, but how I don't know how to move on.

Many Thanks.

>>

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)

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