How many solutions does $z + i|z|= 0$ have? I figured out that it can be transformed in $|z| = iz$ and using the trigonometric method I get:
$|z| = |z|(\cos(x) + i\sin(x))(\cos(\pi/2) + i\sin (\pi/2))$
which becomes
$|z| = |z|(\sin(x) - i\cos(x))$
I delete $|z|$ from both sides and get
$1 = \sin(x) - i\cos(x)$
But don't know how to continue...
PS: I am open to other types of solutions, if more elegant.
Thank you
 A: $|z|$ is a positive number.  So $iz$ must be an imaginary number with positive imaginary part.  That implies $z$ is on the imaginary axis, with a sign opposite that of $i$.  That's all that is needed.  In other words $z=ci$ is a solution for every $c<0$.  You get as many solutions as there are negative numbers $c$.
A: Write $z = x + iy$ with $x, y \in \mathbb R$. Then $x + i(y + \sqrt{x^2 + y^2}) = 0$. Since $x, y \in \mathbb R$, we find $x = 0$ and $y = -\sqrt{x^2 + y^2} = |y|$. These conditions are fulfilled for all $y \le 0$, so the set of solutions is $\{ iy \mid y \le 0\}$.
A: Clearly, $z=-i|z|$ is purely imaginary $=iy$ (say) where $y$ is real.
So, $iy+i|y|=0\implies |y|=-y\implies y\le 0$
A: Infinitely many, $z = -i\alpha$ is a solution for every $\alpha \geq 0$ ($\alpha \in \mathbb{R}$, of course).
A: Let $z=r e^{i \theta}$, this gives the equation $r ( e^{i \theta} + i) = 0$, with $r \geq 0$.
Hence either $r=0$ or $\theta = 2 \pi k -\frac{\pi}{2}$, for some integer $k$.
It follows that the solutions are  $z = -r i$, with $r \geq0$.
