# Equation in c

• July 28th 2009, 08:59 AM
dhiab
Equation in c
• July 28th 2009, 09:15 AM
red_dog
$(z-1)^n=(\overline{z}+1)^n\Rightarrow\left(\frac{z-1}{\overline{z}+1}\right)^n=1$

Apply the modulus to both members. Then
$\left|\frac{z-1}{\overline{z}+1}\right|=1\Rightarrow|z-1|=|\overline{z}+1|$

Let $z=x+yi$

$(x-1)^2+y^2=(x+1)^2+y^2\Rightarrow x=0, \ y\in\mathbb{R}$

So, $z=yi$

Replace y in the equation: $(-1+yi)^n=(1-yi)^n\Rightarrow(-1+yi)^n=(-1)^n(-1+yi)^n$

If n is even the equation is verified for every real y.

If n is odd the equation has no solution.