Need help solving this equation with complex numbers

Question

$$(z^3 + 1)^3 = 1$$

where $z$ is an element of the complex number system.

Can someone show me the most efficient way of finding all the solutions for $z$ here and also if possible please demonstrate how both sides of the equation can be converted into polar coordinates and then the equation be solved?

Thanks a lot!!

Answer 1

The one property you need to know is that :

• $Z^n=r e^{i\theta}$ if and only if $|Z|=\sqrt[n]{r}$ and $n\times\arg(Z) \equiv \theta[2\pi]$
• $1=e^0$ and $-1 = e^{i\pi}$.

Manipulating those two rules lead you to the solutionS.

Answer 2

As $\displaystyle(z^3+1)^3=1=\cos0+i\sin 0,$

using this, $\displaystyle z^3+1=\cos\dfrac{(2k\pi+0)}3+i\sin\dfrac{(2k\pi+0)}3$ where $k=0,1,2$

$\displaystyle\implies z^3=\cos\dfrac{(2k\pi+0)}3+i\sin\dfrac{(2k\pi+0)}3-1$

Using Double-Angle Formulas,

$\displaystyle z^3=2i\sin\frac{k\pi}3\cos\frac{k\pi}3-2\sin^2\frac{k\pi}3$

$\displaystyle=2i\sin\frac{k\pi}3\left(\cos\frac{k\pi}3+i\sin\frac{k\pi}3\right)$

$\displaystyle=2\sin\frac{k\pi}3\left(\cos\frac\pi2+i\sin\frac\pi2\right)\left(\cos\frac{k\pi}3+i\sin\frac{k\pi}3\right)$ as $\displaystyle\cos\frac\pi2+i\sin\frac\pi2=i$

$\displaystyle\implies z^3=2\sin\frac{k\pi}3\left[\cos\left(\frac\pi2+\frac{k\pi}3\right)+i\sin\left(\frac\pi2+\frac{k\pi}3\right)\right]$

Again apply this