Solve $z^2+i\bar{z} = 0$ Need to solve:
$$z^2+i\bar{z} = 0$$
I have tried to use the same method for the other exercise here: Solve $z^2+iz=0$
but I do not know how to manage the $\bar{z}$
Any help?
 A: $$z^2 = - i\bar z$$
so taking absolute value
$$|z|^2 = |\bar z| = |z|$$
so $|z|$ is either $0$ or $1$. If it's $0$ then $z=0$.If it's 1, then we multiply the original equation by $z$
$$z^3 + i z \bar z = z^3 + i |z|^2=  z^3 + i= 0;$$
so 
$$(iz)^3 = 1$$
so the solutions are $z = 0, -i w_k$ where $w_k$ are the cubic roots of 1.
A: let $z = a+bi$, $\overline{z} = a-bi$
$(a+bi)^2+i(a-bi)=0$
$(a^2-b^2+b)+(2ab+a)i = 0$
$2ab+a = 0$ and $a^2-b^2+b =0$
1)if $a= 0$, $b = 0$ or $b = 1$.
2)if $b = -1/2$, $a = \frac{\sqrt3}2$
A: Polar coordinates...
$$
z = re^{i\theta},\qquad z^2=r^2e^{i2\theta},\qquad \overline{z} = re^{-i\theta}
\\
z^2+i\overline{z}=0
\\
z^2=-i\overline{z}
\\
r^2e^{i2\theta}=re^{-i\frac{\pi}{2}-i\theta}
$$
so $r=0$ or
$r=1$ and $2\theta=-\frac{\pi}{2}-\theta\pmod{2\pi}$.
Then
$3\theta=-\frac{\pi}{2}\pmod{2\pi}$ so
$\theta=-\frac{\pi}{6}\pmod{\frac{2\pi}{3}}$
and we end up with three more solutions (in addition to $0$):
$$
z=e^{-i\pi/6} = \frac{\sqrt{3}}{2}-\frac{1}{2}\;i
\\
z=e^{-i5\pi/2} = -\frac{\sqrt{3}}{2} -\frac{1}{2}\;i
\\
z=e^{i\pi/2} = i
$$
A: Rewrite $z$ as $x+iy$:
$(x+iy)^2 + i(x - iy) = 0$ 
$x^2 - y^2 +2ixy + ix + y = 0$
Both the real and imaginary parts must equal zero:
$x^2 - y^2 + y = 0$ and 
$2xy +x = 0$ 
