Show that if $(z+1)^{100} = (z-1)^{100}$, then $z$ is purely imaginary

Question

Let $z$ be a complex number satisfying $$(z+1)^{100} = (z-1)^{100}$$ Show that $z$ is purely imaginary, i.e. that $\Re(z) = 0$.

Rearrange to

$$\left(\frac{z+1}{z-1}\right)^{100} = 1$$

I tried using $z = x+iy$ and trying to multiply numerator and denominator by the conjugate, but I hit a roadblock. I also tried substituting $1 = -e^{i\pi}$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!

Answer 1

Hint $$|z+1|^2=|z-1|^2 \\ \left(z+1\right) \overline{\left(z+1\right)}=\left(z-1\right) \overline{\left(z-1\right)} \\ z \bar{z}+z+\bar{z}+1=z\bar{z}-z-\bar{z}+1 \\ \bar{z}=-z $$

Answer 2

Take absolute values, $|z+1|^{100}=|z-1|^{100}$. Absolute values are always non-negative real numbers, for which root extractions are always defined and one-to-one functions; so take $100^{th}$ roots to get $|z+1|=|z-1|$. Oops, I have an urgent call and need you to continue from there ;-) .

Answer 3

Note that $|z+1|=|z-1|$

Let $z=x+iy$

Therefore

$\sqrt {(x+1)^2+y^2}=\sqrt {(x-1)^2+y^2}$

$(x+1)^2+y^2=(x-1)^2+y^2$

$x^2+2x+1=x^2-2x+1$

$4x=0$

$\therefore x=0$

As we have found out, $z=x+iy\implies z=iy$ therefore yes, it is purely imaginary.