complex numbers quadratic equation question

Question

how to solve $z^2 +3|z| = 0 , z$ complex ? treating the complex number as $a+bi $ or anything similar didnt help much...also solving like simple algebric equations also didnt prove effective and didnt solve this the right way. are there other methods to solution?

Answer 1

Using $z=re^{it}$, you have $$r^2e^{2it}+3r=0$$ $$r(re^{2it}+3)=0\ .$$ Hence $r=0$. i.e. z=0, and $re^{2it}=-3$, i.e r=3 and $e^{2it}=-1$, i.e. $e^{it}=\pm i$.

The three solutions are thus $z=0$ and $z=\pm 3i$.

Answer 2

HINT:

$z^2=-3|z|$ which is real $\implies z$ is either purely imaginary or purely real as demonstrated below

$z=a+ib\implies a^2-b^2+2abi+2\sqrt{a^2+b^2}=0$

Equating the imaginary parts, $ab=0$

Answer 3

If we use exponential representation $z=\rho e^{i \theta}$ then $$\rho^2 e^{2i \theta}+3\rho=0 \Rightarrow \rho=0 \;\;\text{or}\;\; \rho e^{2i\theta}+3=0 \\ \rho e^{2i\theta}+3=0 \Leftrightarrow \rho=-3e^{-2i\theta} \Leftrightarrow -2\theta =\pi(2n+1) \;\;\text{and}\;\;\rho=3$$

Answer 4

Here is another solution.

$$z^2=-3|z|,$$ taking the norm gives

$$|z|^2=3|z|$$ so $|z|=3$ or $|z|=0$

If $|z|=3$ we have $z^2=-9$ and $z=\pm 3i$ so the solutions are $z=0, \pm 3i$