How to solve the equation $2z^{2}+i=-2$? 
How to solve the equation $2z^{2}+i=-2$?

Assuming that $k=-1-i/2$ we have $|k|=\sqrt{5}/2$ and 
$$
\cos \phi=(-2\sqrt5)/5,\quad \sin \phi= -\sqrt5/5.
$$ 
How should I find $\phi$?
 A: Hint: With $$z=x+iy$$ we get $$2(x^2-y^2+1)+i(4xy+1)=0$$
It must be $$x^2-y^2+1=0$$ and $$4xy+1=0$$
A: We have that
$$2z^{2}+i=-2\iff z^2=-1-\frac12 i=\frac{\sqrt 5}2 \left(\cos \theta+i\sin \theta\right)$$
with $\theta = \pi +\arctan \left(\frac12\right)$ (i.e. III quadrant), then refer to De Moivre's formula.
A: You won't find the exact value of $\phi$.
Here is a general method for solving $z^2=p+qi$, where $p$ and $q$ are real.
If $z=a+bi$ where $a$ and $b$ are real numbers, then $z^2=a^2-b^2+2abi=p+qi$, hence $p=a^2-b^2$ and $q=2ab$. You can also notice that $\vert z\vert^2=a^2+b^2=\sqrt{p^2+q^2}$, so you can find $a^2$ and $b^2$, and since you know the sign of $ab$, you can find two couples of solutions $(a,b)$.
In your case, $z^2=-1-\dfrac i2$, so if $z=a+bi$, then $$a^2-b^2=-1$$ $$2ab=-\dfrac 12$$ and $$a^2+b^2=\sqrt{\left(-1\right)^2+\left(-\dfrac 12\right)^2}=\dfrac{\sqrt 5}2\,.$$
Hence $$a^2=\dfrac{\dfrac{\sqrt 5}2-1}2=\dfrac{\sqrt{5}-2}4$$ and $$b^2=\dfrac{\dfrac{\sqrt 5}2+1}2=\dfrac{\sqrt{5}+2}4$$
Since $ab<0$, $a$ and $b$ have opposite signs, and we find that $$(a,b)=\left(\pm\dfrac{\sqrt{\sqrt{5}-2}}2,\mp\dfrac{\sqrt{\sqrt{5}+2}}2\right)$$ and so 
$$z=\pm\dfrac{\sqrt{\sqrt{5}-2}-i\sqrt{\sqrt{5}+2}}2$$
