Solve complex equation $z^3=\left(\sqrt3+i\right)^9$

Question

Solve complex equation $z^3=\left(\sqrt3+i\right)^9$

$|z|=\sqrt{3+1}=2\\\sin \phi=\frac{1}{2}\\\cos\phi=\frac{\sqrt3}{2},\phi=\frac{\pi}{6}\\z_0=\sqrt[3]{2}\left(\cos\frac{\frac{\pi}{6}}{3}+i\sin\frac{\frac{\pi}{6}}{3}\right)=\sqrt[3]2\left(\cos\frac{\pi}{18}+i\sin\frac{\pi}{18}\right)$
Wolfram say $z_0=8i$ what I do wrong?

Answer 1

I think you are taking RHS as $(\sqrt 3+i)^{3}$ instead of $(\sqrt 3+i)^{9}$

Also the given cubic has three solutions. $8i$ is just one of them.

Answer 2

Here $z=(\sqrt{3}+i)^3$ is a solution of the given equation. Now expanding the right hand side you get 8i. As pointed by K R Murthy, it is one of the three solutions.

Answer 3

Hint: use the exponential form of complex numbers: $$\sqrt 3+i=2\biggl(\frac{\sqrt 3}2+\frac 12i\biggr)=2\,\mathrm e^{\tfrac{i\pi}6},$$ so, if you set $z=r\mathrm e^{i\theta}$ the equation is $$z^3=r^3\mathrm e^{3i\theta}=2^9\mathrm e^{\tfrac{i3\pi}2}\iff \dotsm$$

Answer 4

Solving cubic equations in this form is very easy. First find one root. Then multiply by the complex conjugate roots of one to find the other two.

The first root can be easily found using $z_0 = {(\sqrt 3 + i)}^3 = 8i$ (after simplifying with the Binomial theorem).

Then to find the other two roots just take $z_1= \omega z_0, z_2 = \omega^2 z_0$ where $\omega, \omega^2$ are either one or the other (doesn't matter which) of the pair of complex conjugate roots of one, namely $-\frac 12 \pm \frac{\sqrt 3}{2}$. You will note that $\omega^3= (\omega^2)^3 = 1$.

Answer 5

Note

$$z^3=(\sqrt3+i)^9 = (8i)^3 = (8e^{i\frac\pi2})^3=8^3e^{i\frac{3\pi}2+i2\pi k}$$

Thus, the solutions are

$$z = (8^3e^{i\frac{3\pi}2+i2\pi k})^{\frac13}= 8 e^{i\frac{\pi}2+i\frac{2\pi k}3}, \>\>\>k = 0,1,2$$

or $z = 8i, \> \pm 4\sqrt3 -4 i$.

Answer 6

$$z^3-(\sqrt3+i)^9=z^3-(2(\cos30^{\circ}+i\sin30^{\circ}))^3=z^3-8(\cos90^{\circ}+i\sin90^{\circ})=$$ $$=z^3-8i=z^3+(2i)^3=(z+2i)(z^2-2iz-4)=(z+2i)((z-i)^2-3)=$$ $$=(z+2i)(z-i-\sqrt3)(z-i+\sqrt3).$$ Can you end it now?