Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x+8y+48
\end{align}$$
Is it okay? Thank you

Here's another approach altogether:
$$\begin{align}
|z-4i|=2|z+4|
&\iff\left|z-4i\over z+4 \right|=2\\
&\iff{z-4i\over z+4 }=2e^{i\theta}\quad\text{for some }\theta\in\mathbb{R}\\
&\iff z={8e^{i\theta}+4i\over1-2e^{i\theta}}
\end{align}$$
