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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?

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?


marked as duplicate by user91500, N. F. Taussig, muaddib, Ken, Jyrki Lahtonen Jul 28 '15 at 13:36

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  • 6
    $\begingroup$ $z^3 = -\bar z$; so, $\rho =0$ (and it's end) or $\rho = 1$; could you proceed? $\endgroup$ – Michael Galuza Jul 21 '15 at 13:53
  • $\begingroup$ @pbs, it's too long for comment. $\endgroup$ – Michael Galuza Jul 21 '15 at 14:06
  • $\begingroup$ You started right, just no need to use the Euler's formula to go to trigonometric form. You actually want to stay in exponential form to avoid addition and subtraction (it's already factored - no polynomials, no trig functions, no factorization needed, just write the solution in one step). $\endgroup$ – orion Jul 21 '15 at 14:10
  • $\begingroup$ @MichaelGaluza how did you get $\rho = 0 or \rho = 1$? $\endgroup$ – Christian Giupponi Jul 21 '15 at 14:19
  • 1
    $\begingroup$ Complex numbers are equal if their modules are equal. You have $\rho^3 = \rho$; so, $\rho=0,1,-1$, but $\rho\ge0$. $\endgroup$ – Michael Galuza Jul 21 '15 at 14:22

$$z^3+\bar z=0 \Rightarrow z^3=-\bar z$$ 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 $z=\cos(\theta)+i \sin(\theta)$.


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 $$z=\pm \frac{\sqrt2}{2}\pm \frac{\sqrt2}{2}i$$

  • 1
    $\begingroup$ Would it not be nicer to write $\theta \in \{ \pm \pi/4, \pm 3 \pi/4 \}$? Anyway - vote up! $\endgroup$ – johannesvalks Jul 21 '15 at 14:53
  • $\begingroup$ Fixed, thanks for the tip. $\endgroup$ – orion Jul 21 '15 at 15:00
  • $\begingroup$ Exactly the way I would have approached! Well done. +1 $\endgroup$ – Mark Viola Jul 21 '15 at 15:15

$z=0$ 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 $-1$.


$$z^3+\overline z=0$$

But $z=a+bi$, $\overline z = a-bi$ so:


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 $0$.

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

The other solution is then $$z=0$$


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:

$$\text{Real:} \quad a+a^3-3ab^2 = 0$$


$$\text{Imaginary:} \quad 3a^2b - b - b^3 = 0$$

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.

  • $\begingroup$ You may want to check this tutorial before answering questions: meta.math.stackexchange.com/q/5020/137035 $\endgroup$ – user137035 Jul 21 '15 at 14:17
  • $\begingroup$ @i.ozturk thank you so much I was looking for those shortcuts $\endgroup$ – Robert Melikyan Jul 22 '15 at 7:43

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?

  • $\begingroup$ These are a bit harder than linear equations to solve. $\endgroup$ – robjohn Jul 21 '15 at 16:18
  • $\begingroup$ @robjohn indeed, but not much harder and i wanted the OP to do at least something by himself $\endgroup$ – gt6989b Jul 21 '15 at 16:30

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 $$ \frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ -\frac{\sqrt2}2+i\,\frac{\sqrt2}2,\ \ \frac{\sqrt2}2-i\,\frac{\sqrt2}2,\ \ \text{ and }-\frac{\sqrt2}2-i\,\frac{\sqrt2}2. $$

  • $\begingroup$ we have an equation of order three, so we have exactly $3$ solutions...I am counting $5$ in your case, this tells me there has to be an error somewhere, for example $\frac{\sqrt2}2-i\,\frac{\sqrt2}2$ is not a solution $\endgroup$ – user190080 Jul 21 '15 at 14:29
  • $\begingroup$ I am wrong, actually this is a solution! sorry for bothering...just need to sort out my mistake $\endgroup$ – user190080 Jul 21 '15 at 14:46
  • $\begingroup$ It is more or less clear that there should be 5 solutions. One is $z=0$. After you sort that one out, you get $z^4+1=0$, which is a degree 4 polynomial equation and then four more solutions should be expected. $\endgroup$ – Martin Argerami Jul 21 '15 at 15:36
  • $\begingroup$ Yes, you're completely right. I wanted to use the FTA right away but was misguided since the stated term is actually no polynomial in $z$, it is after your manipulation though. $\endgroup$ – user190080 Jul 21 '15 at 16:03

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 $$ z_1 = \frac{-1+\mathbf{i}}{\sqrt{2}}\\ z_2 = \frac{-1-\mathbf{i}}{\sqrt{2}}\\ z_3 = \frac{1-\mathbf{i}}{\sqrt{2}}\\ z_4 = \frac{1+\mathbf{i}}{\sqrt{2}} $$



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.


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