Why $z+i|z|=0$ has infinitely many solutions?

Question

Using the trigonometric form, the equation becomes: $\phi(\cos(\theta)+i\sin(\theta))=-\phi(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$. Since $\phi$ must be equal on both the sides, $\phi=0.$ It should be then one solution, which is $(0,0).$

Answer 1

No, that's incorrect: from $ab=0$ you can deduce that $a=0$ or $b=0$ and you're forgetting the second possibility.

Surely $z=0$ is a solution. Now let's assume $z\ne0$ and look for further solutions, if any. If we write $z/|z|=u$, we have that $|u|=1$ and then the equation becomes $$ |z|u+i|z|=0 $$ and thus $u=-i$, as $|z|\ne0$; this satisfies the requirement, because $\lvert-i\rvert=1$. Moreover we see that the modulus of $z$ can be chosen arbitrarily. Indeed, if $r$ is a positive real number, then $z=-ri$ satisfies the equation.

Answer 2

Using $z=re^{i\theta}$ where $r\ge0$ gives $$re^{i\theta}+ir=0$$ $$r(e^{i\theta}+i)=0$$ So either $r=0\implies z=0$ or $e^{i\theta}=-i\implies z=-ri$.