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
 A: If $(z-2018)^{2n}+(z+2018)^{2n}=0$, then $|z-2018|^2=|z+2018|^2$.
It is your turn to show:
$|z-2018|^2=|z+2018|^2 \iff z+ \overline{z}=0 \iff Re(z)=0.$
A: 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.
A: We have $$\left(\dfrac{z-2018}{z+2018}\right)^{2n}=-1=e^{(2m+1)\pi i}$$ where $m$ is any integer
$$\dfrac{z-2018}{z+2018}=e^{(2m+1)\pi i/2n}$$ where $0\le m<2n$
$$\dfrac z{2018}=\dfrac{1+e^{(2m+1)\pi i/2n}}{1-e^{(2m+1)\pi i/2n}}=-\dfrac{e^{(2m+1)\pi i/4n}+e^{-(2m+1)\pi i/4n}}{e^{(2m+1)\pi i/4n}-e^{-(2m+1)\pi i/4n}}=-\dfrac{2\cos\dfrac{(2m+1)\pi}{4n}}{2i\sin\dfrac{(2m+1)\pi}{4n}}=?$$
using How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i*\sin(\varphi)$?
A: Then simplifying the equation gives us: $$\sum_{k=0}^{2n} \binom{2n}{k} z^{\color{red}{2}n-k} \left[(2018)^k +(-2018)^k \right] =0 \tag 1$$ 
Note that when $k$ is odd, the bracket is always zero, so no problem. But, when $k$ is even, $(1)$ becomes: $$\sum_{k=0}^{n} \binom{2n}{2k} z^{2n-2k} \left[2\times (2018)^{2k} \right] =0$$
Can you take it from here? 
