Complex equation $z^2 + i\overline{z} = 0$

Question

I need to calculate the trigonometric form of the complex solution and then show the algebric form.

$$z^2 + i\overline{z} = 0$$

As far as I know when I have in the equation the $\overline{z}$ the best way to solve it is using: $z= a+ib$ so:

$$(a+ib)^2+i(a-ib)=0$$ $$a^2-b^2+2aib + ia + b = 0$$

Now I separate the real part from the other:

$$\begin{cases}a^2-b^2+b = 0 \\ 2ab + a = 0\end{cases}$$

So I solved the second:

$$2ab+a=0$$ $$a(2b+1)=0$$ $$a=0 \vee b=-\frac{1}{2}$$

Now I tried to solve the first with $a=0$ and I get: $b=0 \vee b=1$

So I have 4 solutions: $a=0, b=-1/2, b =0, b=1$

Is this correct? How should I procede?

EDIT: I have corrected the issue with $\overline{z}$

Answer 1

So I have 4 solutions: $a=0, b=-1/2, b =0, b=1$ Is this correct?

If $a=0$, then $b=0$ or $b=1$. (you are correct here)

If $b=-1/2$, then from the first equation, $a^2=b^2-b=3/4\implies a=\pm\frac{\sqrt 3}{2}$ (you have not considered this case)

Therefore, $$(a,b)=(0,0), (0,1),\left(\frac{\sqrt 3}{2},-\frac 12\right),\left(-\frac{\sqrt 3}{2},-\frac 12\right)$$

Answer 2

An alternative approach:

$$z^2 + i\overline{z} = 0 \implies z^2=-i\overline{z}\implies \vert z \vert^2=\vert -i\overline{z} \vert =\vert z \vert$$

So $\vert z \vert =0 \text{ or } 1$. $\vert z \vert = 0$ gives $z=0$ so we take that solution, and consider the case $\vert z \vert=1$ going forward.

For unit complex numbers, $\overline{z}=z^{-1}$, so we're now solving:

$$z^2 + iz^{-1} = 0\implies z^3=-i=\exp(-i\pi/2)$$

Finding the cube roots of $-i$ using the polar form, we arrive at our final solution set as:

$$\{0,e^{-i\pi/6},e^{i\pi/2}, e^{7i\pi/6} \}$$