Find $z \in \mathbb{C}$ such that |$z+3i| = 3|z|$

Question

I thought I could transform it to

$|z| + |3i| = 3|z|$

$|3i| = 2|z|$

$1.5 = \sqrt{a^2 + b^2}$

But the solution is {$ z = a+bi : a^2 + (b - 3/8)^2 = 81/64 $}

How do I get there?

Answer 1

If $z=a+bi$, with $a,b\in\mathbb R$, then\begin{align}\lvert z+3i\rvert=3\lvert z\rvert&\iff\lvert z+3i\rvert^2=9\lvert z\rvert^2\\&\iff a^3+(b+3)^2=9a^2+9b^2\\&\iff8a^2+8b^2-6b=9\\&\iff a^2+b^2-\frac34b=\frac98\\&\iff a^2+\left(b-\frac38\right)^2=\frac98+\frac9{64}=\frac{81}{64}.\end{align}

Answer 2

$$ |z+3i|^2 = 9|z|^2,$$

$$ x^2 + (y+3)^2 = 9 x^2 + 9 y^2, $$

$$ 8 x^2 + 8 y^2 -6y = 9, $$

$$ 8 x^2 + 8 \left(y-\frac{3}{8}\right)^2 = \frac{63}{8}.$$

This is a circle.

Answer 3

Given |$z+3i| = 3|z|$, we have

$$(z+3i)(z-3i) = 3z\bar z\implies z\bar z-\frac{3i}8(z-\bar z)=\frac98$$

Then, rewrite the equation as

$$\bigg|z - \frac{3i}8\bigg|^2 = \left(\frac98\right)^2$$

which is a circle of center $\frac{3i}8$ and radius $\frac98$.