solve: $z^2+i=|z|$ Solve this: $$z^2+i=|z|$$
I started playing the square then I high again at the square!
$$(z^2+i)^2=|z|^2$$
but after that the calculations become many! there is a better way to solve it?
 A: Hint: $|z|^4 = |z^2|^2=\left||z|-i\right|^2 = |z|^2+1$.
A: HINT:
Let $z=r\cos(\theta+i\sin\theta)$
$\implies |z|=r$ and $z^2=\{r\cos(\theta+i\sin\theta)\}^2=r^2(\cos2\theta+i\sin2\theta)$
So we have, $r^2(\cos2\theta+i\sin2\theta)+i=r$
Equating the real & the imaginary parts 
$r^2\cos2\theta=r\ \ \ \ (1)$ and $r^2\sin2\theta+1=0 \ \ \ \ (2)$ 
From  $(1),$ either $r=0$ or $r\cos2\theta=1$
$r=0$ does not satisfy $(2)\implies r\cos2\theta=1\iff \cos2\theta=\frac1r$
From $(2)\sin2\theta=-\frac1{r^2}=-\cos^22\theta=\sin^22\theta-1$
Solve for $\theta$ from $\sin^22\theta-\sin2\theta-1=0$
A: $$z^2 + i = |z|$$
$$(r \cos(\theta) + r i \sin(\theta))^2 + i = r$$
$$r^2\cos(2\theta) + r^2 i \sin(2\theta) + i = r$$
separate real and imaginary parts
$$r^2 \cos(2\theta) = r$$
$$r^2 \sin(2\theta) = -1$$
Squaring both sides introduces new solutions, so you'll have to double check the final result.
$$r^4 \cos(2\theta)^2 = r^2$$
$$r^4 \sin(2\theta)^2 = 1$$
Add the equations.
$$r^4 = r^2 + 1$$
Now you have a quadratic equation in r^2.  It will have a few extra answers because of the squaring, so double check the result.
