Question

$23^*$. The number of solutions of $z^3 + \overline{z} = 0$ is

Answer 1

5

Answer 2

To find the number of solutions of the equation $z^3 + \overline{z} = 0$, we can use the polar form of a complex number.

Let $z = re^{i\theta}$, where $r = |z| \ge 0$ is the modulus and $\theta = \arg(z)$ is the argument. Then $\overline{z} = re^{-i\theta}$.

Substitute these into the given equation: $(re^{i\theta})^3 + re^{-i\theta} = 0$ $r^3 e^{i3\theta} + re^{-i\theta} = 0$

Factor out $r$: $r(r^2 e^{i3\theta} + e^{-i\theta}) = 0$

This equation holds if either $r=0$ or $(r^2 e^{i3\theta} + e^{-i\theta}) = 0$.

Case 1: $r=0$

If $r=0$, then $z=0$. Substitute $z=0$ into the original equation: $0^3 + \overline{0} = 0 + 0 = 0$. So, $z=0$ is one solution.

Case 2: $r \ne 0$

If $r \ne 0$, then we must have $r^2 e^{i3\theta} + e^{-i\theta} = 0$. This can be rewritten as: $r^2 e^{i3\theta} = -e^{-i\theta}$

We know that $-1 = e^{i\pi}$. So, $r^2 e^{i3\theta} = e^{i\pi} e^{-i\theta}$ $r^2 e^{i3\theta} = e^{i(\pi - \theta)}$

For two complex numbers in polar form to be equal, their moduli must be equal and their arguments must be equal (up to a multiple of $2\pi$).

Equating moduli: $|r^2 e^{i3\theta}| = |e^{i(\pi - \theta)}|$ $r^2 |e^{i3\theta}| = |e^{i(\pi - \theta)}|$ Since $|e^{ix}| = 1$ for any real $x$: $r^2 \cdot 1 = 1 \cdot 1$ $r^2 = 1$ Since $r$ is a non-negative real number, $r=1$.

Equating arguments: $3\theta = (\pi - \theta) + 2k\pi$, where $k$ is an integer. $4\theta = \pi + 2k\pi$ $\theta = \frac{(2k+1)\pi}{4}$

We need to find distinct values of $\theta$ in the interval $[0, 2\pi)$. For $k=0$: $\theta = \frac{\pi}{4}$ For $k=1$: $\theta = \frac{3\pi}{4}$ For $k=2$: $\theta = \frac{5\pi}{4}$ For $k=3$: $\theta = \frac{7\pi}{4}$ For $k=4$: $\theta = \frac{9\pi}{4} = 2\pi + \frac{\pi}{4}$, which is equivalent to $\frac{\pi}{4}$. So, we have found all distinct values.

These 4 distinct values of $\theta$ (with $r=1$) give 4 distinct solutions:

1. $z_1 = 1 \cdot e^{i\pi/4} = \cos(\pi/4) + i\sin(\pi/4) = \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$
2. $z_2 = 1 \cdot e^{i3\pi/4} = \cos(3\pi/4) + i\sin(3\pi/4) = -\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}$
3. $z_3 = 1 \cdot e^{i5\pi/4} = \cos(5\pi/4) + i\sin(5\pi/4) = -\frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$
4. $z_4 = 1 \cdot e^{i7\pi/4} = \cos(7\pi/4) + i\sin(7\pi/4) = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}}$

Combining both cases, we have 1 solution from Case 1 ($z=0$) and 4 solutions from Case 2. Total number of solutions = $1 + 4 = 5$.

The equation $z^3 + \overline{z} = 0$ is solved by considering two cases based on the modulus of $z$.

1. If $|z|=0$, then $z=0$, which is a solution.
2. If $|z| \ne 0$, let $z=re^{i\theta}$. Taking the modulus of $z^3 = -\overline{z}$ yields $|z|^3 = |-\overline{z}| = |\overline{z}| = |z|$. This simplifies to $r^3=r$, which gives $r(r^2-1)=0$. Since $r \ne 0$, we get $r^2=1$, so $r=1$.

For $|z|=1$, we have $\overline{z} = \frac{1}{z}$. Substituting this into the original equation gives $z^3 + \frac{1}{z} = 0$. Multiplying by $z$ (since $z \ne 0$), we get $z^4+1=0$, or $z^4=-1$. The solutions are the four distinct fourth roots of $-1$. These are $e^{i\pi/4}, e^{i3\pi/4}, e^{i5\pi/4}, e^{i7\pi/4}$. Adding the $z=0$ solution, the total number of solutions is $1+4=5$.