Find all $z \in \mathbb{C}$ that satisfy $z^3 = −2(1 + i\sqrt{3})\bar{z}$

Question

Find all $z \in \mathbb{C}$ that satisfy $z^3 = −2(1 + i\sqrt{3})\bar{z}$. You must express your answers in the standard form.

So far, I'm thinking of writing $z = a + bi$, but then I have to expand the bracket, and I feel like it is way too much work and that there has to be a simpler way to solve this.

Any suggestions would be appreciated, thanks $:)$

Answer 1

When dealing with multiplication of complex numbers, it is generally easier to work with polar form. If you let $z = re^{i\theta}$ then you obtain

$$r^3e^{3i\theta} = -2(1+i\sqrt{3})re^{-i\theta}.$$

The next step is to write $-2(1+i\sqrt{3})$ is polar form as follows:

$$-2(1+i\sqrt{3}) = 4\left(-\frac{1}{2}-i\frac{\sqrt{3}}{2}\right) = 4e^{i\frac{4\pi}{3}}.$$

So we have

$$r^3e^{3i\theta} = 4e^{i\frac{4\pi}{3}}re^{-i\theta} = 4re^{i\left(\frac{4\pi}{3}-\theta\right)}.$$

Now use the fact that two complex numbers written in polar form are equal if and only if they have the same modulus and the same argument up to an integer multiple of $2\pi$.

Answer 2

Put $\;z=re^{it}\implies z^3=r^3e^{3ti}\;,\;\;\overline z=re^{-it}\;,\;\;-2-2\sqrt3\,i=4\,e^{4\pi i/3}$

Now you get:

$$r^3e^{3ti}=4\,e^{4\pi i/3}re^{-ti}\iff r^2e^{4ti}=4e^{4\pi i/3}$$

Take it from here, and use periodicity of trigonometric functions...or not, depending on how many solutions you want.

Answer 3

HINT. Let us consider the exponential form of complex number: $$z=\rho e^{i\theta}$$ and express in exponential form $-2-i2\sqrt 3$.