Find All Complex solutions for $z^3+3i\bar{z}=0$

Question

Find All Complex solutions for $z^3+3i \bar z =0$. I tried substituting $z=a+bi$ and simplifying as much as I can and this is what I ended up with: $a^3-2b^2a+3b+i(3ba^2-b^3+3a)=0$.

I just did not understand how do I get the values of $z$ from this equation

Answer 1

Since $z^3 =-3i\bar{z}$ we have $\bar{z}^3 = 3iz$. So pluging $\bar{z} ={iz^3\over 3}$ in second equation we get$${-iz^9\over 27} = 3iz\implies z^9+81z=0$$

So $z_9=0$ and then $$z^8=-81\implies z_k=\sqrt{3}(\cos{(2k-1)\pi\over 8}+i\sin{(2k-1)\pi\over 8}) $$ $k\in\{1,2,...8\}$

Answer 2

Write $z=r e^{i\theta}$; the equation becomes $$r^3 e^{3 i\theta}+3i re^{-i\theta}=0.$$ One immediate solution has $r=0$, i.e., $z=0$. Otherwise, we only need to solve $$r^2 e^{4 i\theta}+3i =0.$$

We rewrite this as $$r^2 e^{4 i\theta}=3 e^{3\pi i/2}.$$ We find that $r=\sqrt{3}$ for the four possibilities $e^{ i\theta}= e^{3\pi i/8}$, $e^{ i\theta}= ie^{3\pi i/8}$, $e^{ i\theta}= -e^{3\pi i/8}$, and $e^{ i\theta}= -ie^{3\pi i/8}$, found from the four fourth roots of unity. These yield $\theta\in(3\pi/8,7\pi/8,11\pi/8,15\pi/8)$.

Answer 3

You let $z=a+bi$ and the system of equations to solve is: $a^3-3ab^2+3b=0$ and $3ba^2-b^3+3a=0$. If you multiply the first equation by $-a$ and the second by $b$ and then add the resulting equations you get $a^4+b^4=6a^2b^2$. Let $m=\frac{b}{a}$. Then, $$m^4-6m^2+1=0.$$ The solution of this equation is: $m=\tan(\frac{3\pi}{8}+\frac{k\pi}{2})$ where $k=0,1,2,3$. Hence, we found the Arguments (angles) of the 4 solutions. The other 4 Arguments can be found by formula $\pi-\theta$ from these Arguments, since $\tan(\pi-\theta)=-\tan\theta$ satisfies the equation too.

To find the magnitudes of them, multiply the first equation by $a$ and the second by $b$ and then add the resulting equations you get $a^2+b^2=\frac{6ab}{b^2-a}$. Then, since $b=ma$, we have $|z|=\sqrt\frac{6m}{m^2-1}$. We find out that $\sqrt\frac{6m}{m^2-1}=\sqrt{3}$ for all $m$ values we found above. I checked by WolframAlpha trying to prove by hand...

The solution $z=0$ could be found by polar method.

Answer 4

This is a commentary on all of the existing answers, rather than a suggested method of solution. It seems fashionable to state that there are $9$ solutions, when in reality there are $5$ solutions $z$ to the equation $$z^3-3i\overline z = 0$$ Writing $z=x+iy$, we obtain the following polynomial equations by taking real and imaginary parts of the given equation: \begin{align*} &x^3-3xy^2+3y = 0 \\ &3x^2y+3x-y^3 = 0 \end{align*} By Bézout's Theorem, the corresponding projective curves have $9$ intersections, counting multiplicity. However, we are seeking real solutions $(x,y)$ in the affine plane, so there could be fewer than $9$ solutions to the equation $z^3-3i\overline z = 0$. By standard polynomial algebra, this system is equivalent to \begin{align*} &9x-8y^7+51y^3 = 0 \\ &64y^9-432y^5+81y=0 \end{align*} Factoring $y$ out of the second equation and performing the substitution $Y=y^4$, we obtain \begin{equation*} 64Y^2-432Y+81 \end{equation*} The roots are $$Y = \frac{27}8 \pm \frac94\sqrt2$$ These are positive real numbers, and so the solutions for $y$ are \begin{equation*} y = 0, \pm\left(\frac{27}8 \pm \frac94\sqrt2\right)^{1/4} \end{equation*} For each $y$, the coordinate $x$ is uniquely determined and real. This means that there are $5$ solutions to the equation, $$z^3-3i\overline z = 0$$