<s>[INST] Solving quadratic complex equation I'm trying to solve this equation:

$z^2+2iz+11-16i=0$
where $z=x+iy$ with $ x,y \in \mathbb{R}$.

I tried it with the quadratic formula $z_{1/2}=- \frac p2 \pm \sqrt{\left(\frac p2\right)^2 -q}$ and get:
$z_{1/2}=-i \pm \sqrt{i^2-11+16i} \\ \Leftrightarrow z_{1/2}=-i \pm \sqrt{12i^2+16i} $
Now I'm stucking. Any ideas how it's going on?
 [/INST] Completing the square might be simpler: 
$z^2+2iz+i^2 = -11+16i+i^2$
$(z+i)^2 = -12+16i$
$z = -i \pm \sqrt{-12+16i}$
$z = -i \pm 2\sqrt{-3+4i}$ (factor out a $4$ from under the squareroot). 
To simplify $\sqrt{-3+4i} = a+bi$, we need to find reals $a,b$ such that $(a+bi)^2 = (a^2-b^2)+(2ab)i = -3+4i$. 
Since the numbers are small, you can try guessing integers for $a,b$. 
Since $a = 1$ and $b = 2$ work, $1+2i$ is a value for $\sqrt{-3+4i}$. Can you continue from here?
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