Magnitude of Complex Numbers Let $\alpha \neq 1$ be a complex number such that the distance from $\alpha^2$ to 1 is twice the distance from $\alpha$ to 1, while the distance from $\alpha^4$ to 1 is four times the distance from $\alpha$ to 1. Enter all possible values of $\alpha,$ separated by commas.
I have no idea how to do this. Can someone help me?
 A: You have
$$
|\alpha^4-1|=2|\alpha^2-1|=4|\alpha-1|.
$$
Let's use bars to denote complex conjugate. Actually we do not need to write $\alpha$ as $a+bi$. We have
$$
|\alpha^4-1|^2=(\alpha^4-1)(\bar\alpha^4-1)=(\alpha^2+1)(\bar\alpha^2+1)(\alpha^2-1)(\bar\alpha^2-1)=|\alpha^2+1|^2|\alpha^2-1|^2,\\
|\alpha^4-1|=|\alpha^2+1||\alpha^2-1|
$$
So, $|\alpha^2+1|=2$. Similarly, $|\alpha+1|=2$
We can rewrite those results as
$$
\alpha^2\bar\alpha^2+\alpha^2+\bar\alpha^2=3\\
\alpha\bar\alpha+\alpha+\bar\alpha=3
$$
From the second equation we get
$$
\bar\alpha=\frac{1-\alpha}{1+\alpha}.
$$
Substituting into the first equation, and factorize a bit, we get
$$
(1+\alpha^2)(3-\alpha)^2-(3-\alpha^2)(1+\alpha)^2=0
$$
This gives us the trivial solution $\alpha=1$. To find other solutions, we can simplify the above equation. We can then obtain
$$
(\alpha-1)(\alpha^3-\alpha^2+3\alpha-3)=(\alpha-1)(\alpha^2+3)(\alpha-1)=0
$$
The other two sultions,
$$
\alpha=-\sqrt{3} i,+\sqrt{3} i
$$
Satisfy the requirements of the question.
A: An alternative method:
Use the first condition to find$$|\alpha^2-1|=2|\alpha-1|\\|\alpha+1|=2\\\alpha=-1+2e^{i\theta}$$
Use the second condition to get $$|\alpha^4-1|=4|\alpha-1|\\|\alpha^3+\alpha^2+\alpha+1|=4\\|(2e^{i\theta}-1)^3+(2e^{i\theta}-1)^2+2e^{i\theta}|=4\\\left|8e^{3i\theta}-12e^{2i\theta}+6e^{i\theta}-1+4e^{2i\theta}-4e^{i\theta}+1+2e^{i\theta}\right|=4\\\left|2e^{2i\theta}-2e^{i\theta}+1\right|=1\\2e^{2i\theta}-2e^{i\theta}+1=e^{i\phi}\\2\cos2\theta-2\cos\theta+1=\cos\phi\in[-1,1]\\2\sin2\theta-2\sin\theta=0$$This second equation tells us $$(4\cos\theta-2)\sin\theta=0$$So $\theta=\pi,\pm\frac\pi3$. If $\theta=\pi$, then the first equation gives $5=\cos\phi$ which is not solvable for real $\phi$, so this is not a solution. Therefore have $\theta=\pm\frac\pi3$. 
$$\alpha=-1+2\left(\frac12\pm i\frac{\sqrt3}2\right)=\pm\sqrt3i$$
