# Complex roots of $z^3 + \bar{z} = 0$

I'm searching for the intricate roots of $z^3 + \bar{z} = 0$ making use of De Moivre.

Some recommended increasing both sides by z first, yet that appears incorrect to me as it would certainly add a root (and also I would not recognize which root was the added ).

I saw that $z=a+bi$ and also there exists $\theta$ such that the trigonometric depiction of $z$ is $\left ( \sqrt{a^2+b^2} \right )\left ( \cos \theta + i \sin \theta \right )$.

It appears that $-\bar{z} = -\left ( \sqrt{a^2+b^2} \right )\left ( \cos (-\theta) + i \sin (-\theta) \right )$

However, my trig is rather corroded and also I'm not fairly certain where to go from below.

Write $z = re^{i\theta}$. Then you are trying to solve $r^3e^{3i\theta} + re^{-i\theta} = 0$, which is the same as $$r^3e^{3i\theta} = -re^{-i\theta}$$ Note that $-1 = e^{i\pi}$, so the above is equivalent to $$r^3e^{3i\theta} = re^{i(\pi - \theta)}$$ Comparing magnitudes, you have $r^3 = r$, which is solved by $r = 0$ and $1$, and comparing arguments you must have $3\theta = \pi - \theta + 2\pi k$ for some integer $k$ (when $r \neq 0$). Thus for some integer $k$ you have $$\theta = {\pi \over 4} + k{\pi \over 2}$$ There are four values of $\theta$ in $[0,2\pi)$ that satisfy this, namely ${\pi \over 4}, {3\pi \over 4}, {5\pi \over 4}$, and ${7\pi \over 4}$. Thus the complex numbers satisfying your original equation are $0, e^{i {\pi \over 4}}, e^{i {3\pi \over 4}}, e^{i {5\pi \over 4}}$, and $e^{i {7\pi \over 4}}$. In rectangular coordinates these are $0$ and $\pm {1 \over \sqrt{2}} \pm {i \over \sqrt{2}}$.

**EDIT** *in view of the comments bellow by JimConant and PeterTaylor. If there is still any error the fault is mine.*

This is an alternative solution to the trigonometric one. We will use the algebraic method. Let $z=x+iy$. We have $$\begin{eqnarray*} 0 &=&z^{3}+\overline{z} \\ 0 &=&\left( x+iy\right) ^{3}+\left( x-iy\right) \\ &=&x^{3}+x-3xy^{2}+i\left( 3x^{2}y-y^{3}-y\right) \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=x(x^{2}+1-3y^{2}) \\ 0=y(y^{2}+1-3x^{2}). \end{array} \right. \end{eqnarray*}\tag{1}$$

One of the roots is $$x_{1}=y_{1}=0.\tag{1a}$$ The remaining *real* roots satisfy the system

$$\begin{eqnarray*} \left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=y^{2}+1-3x^{2} \end{array} \right. &\Leftrightarrow &\left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=\frac{1}{3}+\frac{1}{3}x^{2}+1-3x^{2} \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=x^{2}+1-3y^{2} \\ 0=4-8x^{2} \end{array} \right. \\ &\Leftrightarrow &\left\{ \begin{array}{c} 0=1-2y^{2} \\ 0=1-2x^{2}. \end{array} \right. \end{eqnarray*} \tag{2}$$

The last system means that $$y=\pm x\tag{3}$$ and that

$$\begin{eqnarray*} x &=&\pm\frac{1}{2}\sqrt{2}, \\ y &=&\pm\frac{1}{2}\sqrt{2}. \end{eqnarray*}\tag{3a}$$

Combining the above results, we conclude that the following five complex numbers

$$ z_{1} =0,\tag{4}$$ $$ z_{2} =\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2},\quad z_{3} =-\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2}, \tag{5}$$ $$z_{4} =\frac{1}{2}\sqrt{2}-i\frac{1}{2}\sqrt{2},\quad z_{5} =-\frac{1}{2}\sqrt{2}+i\frac{1}{2}\sqrt{2}, \tag{6}$$

are the solutions of the given equation $$z^{3}+\overline{z}=0.\tag{7}$$

OK, I'm going to take a stab at this.

Given: $z^3 + \bar{z} = 0$

Therefore $z^3 = -\bar{z}$ and $|z^3| = |-\bar{z}|$ .

We have that $|-\bar{z}|=|\bar{z}|=|z|$ therefore $|z^3|=|z|$ and $|z||z||z|=|z|$.

Therefore $|z|=0$ or $|z|=1$.

Case 1: Assume $|z|=0$ then if

$z=a+bi\leftrightarrow|z|=\sqrt{a^2+b^2}=0\leftrightarrow a=0 \wedge b=0\leftrightarrow a+bi=0$

thus $z = 0$.

Case 2: Assume $|z|=1$

Let's multiply by $z$ and we get $z^4 = -z\bar{z}$. We see that $z\bar{z} = |z|^2$ so we get $z^4 = -( |z|^2 )$ so from the above either:

therefore $z^4 = -|z| = -1$

The trigonometric representation of $-1$ is $1*( \cos \pi + i \sin \pi )$ so according to De Moivre:

$z^4 = r^4(\cos 4\theta + i \sin 4\theta ) = 1*( \cos \pi + i \sin \pi )$

These are two complex numbers in trigonometric form so:

$r^4 = 1$ and $4\theta = \pi + 2\pi*k$ or

$r=1$ and $\theta = \frac{\pi + 2\pi*k}{4}$ and each solution has the form:

$z_k = \cos( \frac{\pi + 2\pi*k}{4} ) + i \sin (\frac{\pi + 2\pi*k}{4})$

for $0\leq k \leq 3$.

Which together with Case 1 gives the following values for $z$:

$0,\pm\frac{1+i}{\sqrt{2}},\pm\frac{1-i}{\sqrt{2}}$

That is a comment to the comment "should there be only 3 roots?" in Zarrax answer. Actually, the questions is "why isn´t there 9 roots?". This is the right question to ask since the intersection of the two curves in $\mathbb{R}^2$ $$ x^3-3xy^2 +x = 0$$ $$ 3x^2y-y^3 -y = 0$$ is your solution set (just expand out $z^3+\bar{z}$). Now, the intersection of two degree 3 equations should have $3x3= 9$ solutions (by Bezout´s theorem). The 4 roots we are missing at "infinity" or in the $\mathbb{C}^2$ plane. We could apply the same thing to the function $z^2+z$. As complex polynomial, we should expect 2 roots. As the real system $(x,y) \rightarrow (x^2-y^2+x,2xy+y)$ we expect four real roots.

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