prove the root of $(z-2018)^{2n}+(z+2018)^{2n}=0$ is purely imaginary Prove: if $z \in \mathbb{C}$ satisfies $(z-2018)^{2n}+(z+2018)^{2n}=0$,then $z=bi$ for some $b \in \mathbb{R} (b \neq 0)$.
The method i can think of is use the binomial theorem to get:
$$\sum_{k=0}^{2n} \binom{2n}{k}z^{2n-k}(-2018)^{k}=-\sum_{k=0}^{2n} \binom{2n}{k}z^{2n-k}(2018)^{k} $$
i feel like it must be that the imaginary part of z makes this equality holds, but it seems like there is no way to prove this argument. 
i guess the $2n$ as the exponential must be the key to solve this, but i can't see how

If $(z-2018)^{2n}+(z+2018)^{2n}=0$, then $|z-2018|^2=|z+2018|^2$.
It is your turn to show:
$|z-2018|^2=|z+2018|^2 \iff z+ \overline{z}=0 \iff Re(z)=0.$
