Complex number: Roots 
Solve all the roots of the following equation: $$(z-i)^2(z+i)^2=\frac{1}{4}.$$ Find the set of complex numbers $z$ such that $$\left|\frac{z-3}{z+3}\right|=2.$$

Would anyone mind telling me how to solve the above problems? I really have no idea.

The first one:
$$(z-i)^2(z+i)^2 = \left( (z-i)(z+i) \right)^2 = (z^2 - i^2)^2 = (z^2+1)^2,$$
so
$$z^2+1 = \pm\frac{1}{2}.$$
For the second one, write $z = x + iy$, so $|z-3| = 2|z+3|$, which is equivalent to $|z-3|^2 = 4|z+3|^2$, gives you
$$(x-3)^2 + y^2 = 4(x+3)^2 + 4y^2.$$
