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

Solution 1:

Hint: First try to narrow down the value of $|z|$

Solution 2:

As has been observed by Thomas Andrews, $z=0$ is a root, so we will not be introducing extraneous roots if we multiply by $z$, or by $\overline{z}$.

From the answer by Aryabhata, you should be able to conclude that if $z\ne 0$, then $z$ has norm $1$.

Now it may be simplest to multiply by $\overline{z}$. We get the pleasantly symmetrical equation $z^2+\overline{z}\overline{z}=0$. If $z$ has norm $1$, let $z=\cos\theta+i\sin\theta$. Then use De Moivre's Theorem.

Solution 3:

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 $\pm {1 \over \sqrt{2}} \pm {i \over \sqrt{2}}$.

Solution 4:

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 $$z^{3}+\overline{z}=0.\tag{7}$$

Solution 5:

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.