Question

Find all the complex numbers z for wich conj(z)=z^3

We're suppose to use the exponential form of z, z=|z|*e^(iθ).

Thanks

Answer 1

The conjugate of \(z=|z|e^{i\theta}\) is \(|z|e^{-i\theta}\).

Answer 2

so you want to solve
|z| exp(-i*theta) = z^3 ?

Answer 3

ok take the third root of both sides

Answer 4

z = |z|^(1/3) * exp(-i/3*theta) , up to moduli 2pi/3

Answer 5

that means you can have multiples of 2pi/3

Answer 6

\[\large\begin{align*}re^{-i\theta}&=\left(re^{i\theta}\right)^3\\\\
r^3e^{3i\theta}-re^{-i\theta}&=0\\\\
\color{red}{re^{-i\theta}}\left(r^2e^{4i\theta}-1\right)&=0
\end{align*}\]
The red factor gives the trivial solution, \(z=0\). You can factorize the remaining one further:
\[\large\begin{align*}
r^2e^{4i\theta}-1&=0\\\\
\left(re^{2i\theta}\right)^2-1&=0\\\\
\left(re^{2i\theta}-1\right)\left(re^{2i\theta}+1\right)&=0
\end{align*}\]
If you consider one factor at a time, you'll find that they have the same solutions, \(z=\pm1\) and \(z=\pm i\).