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

Assume $z_0$ is a solution. In that case we have
$$
(z_0-2018)^{2n}+(z_0+2018)^{2n}=0\\
(z_0-2018)^{2n}= -(z_0+2018)^{2n}\\
\left|(z_0-2018)^{2n}\right|= \left|-(z_0+2018)^{2n}\right|\\
\left|z_0-2018\right|^{2n}= \left|z_0+2018\right|^{2n}\\
|z_0-2018| = |z_0 + 2018|
$$
where we can take the $2n$-th root without issues because both absolute values are non-negative real numbers. The last line says that the distance in the complex plane from $z_0$ to $2018$ is the same as the distance from $z_0$ to $-2018$. So $z_0$ must lie on the imaginary axis.
As far as I can see, we only need the exponent to be even in order to exclude $b = 0$ as a possibility.
