Solve $z^2+i\bar{z} = 0$

Question

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?

Answer 1

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 $b = -1/2$, $a = \frac{\sqrt3}2$

Answer 2

$$z^2 = - i\bar z$$ 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.

Answer 3

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

$2xy +x = 0$

Answer 4

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 $0$): $$ z=e^{-i\pi/6} = \frac{\sqrt{3}}{2}-\frac{1}{2}\;i \\ z=e^{-i5\pi/2} = -\frac{\sqrt{3}}{2} -\frac{1}{2}\;i \\ z=e^{i\pi/2} = i $$