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

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

$$left(fracz+1z-1right)^100 = 1$$

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 $1 = -e^ipi$, but that also doesn't seem to get me anywhere. How can I prove this? Any help is appreciated, thank you!

3 Answers
3

Hint
$$|z+1|^2=|z-1|^2 \
left(z+1right) overlineleft(z+1right)=left(z-1right) overlineleft(z-1right) \
z barz+z+barz+1=zbarz-z-barz+1 \
barz=-z
$$

Take absolute values, $|z+1|^100=|z-1|^100|$. Absolute values are always no 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 ;-) .

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

Therefore

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

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

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

$4x=0$

$therefore x=0$

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