Complex number: Roots

Question

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.

Answer 1

For the first question, you have $(z-i)(z +i) = z^2 + 1$. This should help you get there.

For the second, you might want to square both sides and clear the fraction.

Try these things and see if they help.

Answer 2

HINTS:

1. Difference of two squares.

2. We can show with a bit of work that

$$\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}.$$

What might help then is if you know what $|z-a|$ is geometrically.

Answer 3

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

$$(x-3)^2 + y^2 = 4(x+3)^2 + 4y^2.$$