# 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.

12
2022-07-25 20:39:40
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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 $\pm {1 \over \sqrt{2}} \pm {i \over \sqrt{2}}$.

3
2022-07-25 22:31:45
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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 $$z^{3}+\overline{z}=0.\tag{7}$$

3
2022-07-25 22:24:22
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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 $z$:

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

1
2022-07-25 22:22:33
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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.

2
2022-07-25 21:39:11
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