How to solve $z^3 + \overline z = 0$ I need to solve this:
$$z^3 + \overline z = 0$$
how should I manage the 0?
I know that a complex number is in this form: z = a + ib so:
$$z^3 = \rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace$$
$$\overline z = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
but how about the 0?
EDIT:
ok, following some of your comments/answers this is what I have done:
$$z^3 = - \overline z$$
$$\rho^3\lbrace \cos(3\theta) + i \sin (3\theta)\rbrace = \rho\lbrace \cos(-\theta) + i \sin (-\theta)\rbrace$$
So
$$
\begin{Bmatrix}
\rho^3 = \rho\\ 
3\theta = -\theta + 2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho^3 = \rho\\ 
2\theta =  2k\pi
\end{Bmatrix}$$
$$
\begin{Bmatrix}
\rho = 0 or \rho = 1\\ 
\theta =  k\frac{\pi}{2}
\end{Bmatrix}$$
is this the right way?
 A: $$z^3+\bar z=0 \Rightarrow z^3=-\bar z$$
Taking absolute values on bot sides you get
$$|z|^3=|\bar{z}|=|z|$$
thus $|z|=0$ or $|z|=1$.
Case 1: $|z|=0 \Rightarrow z=0$.
Case 2 $|z|=1$. Multiply your original equation by $z$ and use $z \bar{z}=1$. Thus you get
$$z^4=-1$$
which is easy to solve in trig form. Remember that $r=1$ thus your $z=\cos(\theta)+i \sin(\theta)$.
A: Just continue what you started:
$$\rho^3 e^{3i\theta}+\rho e^{-i\theta}=0$$
$$\rho^2=-e^{-4i\theta}$$
As $\rho$ is positive and real, and exponential of an imaginary argument is on a unit circle, you know that the only solution is $\rho=1$ and $e^{-4i\theta}=-1$ meaning
$$\theta\in\lbrace \pm\pi/4, \pm 3\pi/4 \rbrace$$
Back in the cartesian form you get the obvious solution $z=0$ (from $\rho=0$), and also
$$z=\pm \frac{\sqrt2}{2}\pm \frac{\sqrt2}{2}i$$
A: $z=0$ is an obvious solution.
Then multiplying by $z$,
$$z^3=-\bar z\implies z^4=-|z|^2.$$
Taking the modulus, $|z|^4=|z^2|=1$ and $z$ is a fourth root of $-1$.
A: $$z^3+\overline z=0$$
But $z=a+bi$, $\overline z = a-bi$ so:
$$(a+bi)^3+a-bi=a^3-3ab^2+3a^2bi-b^3i+a-bi=i(3a^2b-b^3-b)+(a^3-3ab^2+a)$$
We know that $a+bi=0 \iff a=0 \land b=0$:
$$\begin{cases}
3a^2b-b^3-b=0\\
a^3-3ab^2+a=0
\end{cases}$$
$$\begin{cases}
3a^2-b^2-1=0\\
a^2-3b^2+1=0
\end{cases}$$
$$a^2=3b^2-1 \implies 3a^2-b^2-1=9b^2-3-b^2-1=8b^2-4=0 \implies b=\pm\frac{\sqrt2}{2}$$
$$a^2=3b^2-1=1.5-1=0.5\implies a=b=\pm\frac{\sqrt2}{2}$$
So finally:
$$z=\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2}i$$
or
$$z=-\frac{\sqrt2}{2}\pm\frac{\sqrt2}{2}i$$

To be clear there may be other solutions. I have divided my equations by $a$ and $b$, but each may be equal to $0$.


*

*$a=0$ then $b^3+b=0$


*

*$b=0$ OK

*$b\not=0$ then $b^2+1=0$ and $b\not\in \mathbb{R}$ (but it's a contradiction)


*$b=0$ then $a^3+a=0$, same as above, only $(a;b)=(0;0)$ is valid.


The other solution is then $$z=0$$
A: Why don't you just use $z= a +ib$, so you would get:
$$a - ib + a^3 + 3(ib)(a^2) - 3(a)(b^2) - ib^3 = 0$$
If it equals $0$ then both the real and imaginary parts are equal to $0$, so:
$$\text{Real:} \quad a+a^3-3ab^2 = 0$$
and
$$\text{Imaginary:} \quad 3a^2b - b - b^3 = 0$$
It's pretty easy from here on out. There are a number ways to solve this: you can make them equal to each, other, solve them separately for variables, etc.
A: HINT
If $z = a+bi$ note that
$$
z^3 = a^3 - 3ab^2 + 3a^2bi - b^3i = (a^3 - 3ab^2) + (3a^2b-b^3)i
$$
so
$$
0 = \bar{z} + z^3 = (a^3 - 3ab^2+a) + (3a^2b-b^3-b)i,
$$
which implies you have 2 equations in 2 unknowns.
Can you take it from here?
A: $$z^3+z^*=0.$$
Let $z=x+iy$. Then
$$(x+iy)^3+x-iy=0,$$ so that
$$x+x^3-3xy^2=0,\tag{1}$$ and $$-y+3x^2y-y^3=0.\tag{2}$$
We ignore the trivial solution $x=y=0$ and suppose that $x,y\neq 0$. Then using (1) and (2) we see that $$x^3+x=x\left(1+\frac{y^3+y}{3y}\right).$$ Hence, $$-3xy^2+x\left(1+\frac{y^3+y}{3y}\right)=0.$$ Since $x\neq 0$ this gives $$-3y^2+1+\frac{y^3+y}{3y}=0,$$ leading to $$y(1-2y^2)=0.$$ Therefore, we must have $$y=\pm\frac{1}{\sqrt{2}}.$$ Substituting into $(2)$ gives $$x=\pm\frac{1}{\sqrt{2}}.$$ Therefore,
$$z=x+iy=\pm\frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}},$$ where each $\pm$ is chosen independently of the other.
A: One solution is $z=0$. When $z\ne0$, multiplying by $z$ will not alter the solutions of the equations; so we can look at $z^4+|z|^2=0$. If we write $z=re^{it}$, the equation becomes 
$$
r^4e^{4it}+r^2=0.
$$
As we are assuming $r\ne0$, this reduces to $-r^2e^{4it}=1$. So $r=1$ and $e^{4it}=-1$, that is (since $-1=e^{i\pi}$), $4t=\pi+2k\pi$. Thus
$$
t=\frac14\,\left(\pi+2k\pi\right)=\frac{\pi}4+\frac{k\pi}2,\ \ k=0,1,2,3.
$$
So we get four solutions, from $z=\cos t+ i\sin t$, which are 
$$
\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ -\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ \frac{\sqrt2}2-i\,\frac{\sqrt2}2,\ \ \text{ and }-\frac{\sqrt2}2-i\,\frac{\sqrt2}2.
$$
A: Given
$$
z^3 = - \bar{z}.
$$
First we note that
$$
|z^3| = |-\bar{z}| \Longrightarrow |z|^3 = |z|.
$$
Therefore
$$
z = 0 \vee z = \exp(\zeta \pi \mathbf{i}).
$$
The case $z \ne 0$
We obtain
$$
\exp(3 \zeta \pi \mathbf{i}) = \exp( \pi \mathbf{i} - \zeta \pi \mathbf{i}).
$$
Whence
$$
3 \zeta = 2 k + 1 - \zeta \Longrightarrow \zeta = \frac{1}{4} + k.
$$

The general solution can be written as
$$
z = 0 \vee z = \exp\Big( \big[ \tfrac{1}{4} + k \big] \pi \mathbf{i} \Big)
= \exp\big( \pi \mathbf{i} / 4 \big) \exp\big( k \pi \mathbf{i} \big).
$$
Or as
$$
z_0 = 0 \vee k \in {1,2,3,4} : z_k = \exp\big( \pi \mathbf{i} / 4 \big) \mathbf{i}^k.
$$

Note that
$$
\exp\big( \pi \mathbf{i} / 4 \big) = \frac{1+\mathbf{i}}{\sqrt{2}},
$$
so
$$
z_1 = \frac{-1+\mathbf{i}}{\sqrt{2}}\\
z_2 = \frac{-1-\mathbf{i}}{\sqrt{2}}\\
z_3 = \frac{1-\mathbf{i}}{\sqrt{2}}\\
z_4 = \frac{1+\mathbf{i}}{\sqrt{2}}
$$
