Finding three roots of a complex number if we already know one root If we know that $a+bi$ is one of the forth roots of the complex number $z$, how can we find the other three roots?
 A: If $(a+bi)^4 = z\;$ then $\big(-(a+bi)\big)^4 = z\;$ and $\big(\pm i(a+bi)\big)^4 = z$
A: If you are looking for the fourth roots of zero, they are all zero. 
Suppose now that you are looking for the fourth roots of $a \in \mathbb{C}\setminus\{0\}$, i.e. the solutions of the equation $z^4 = a$. If you know that $z_0$ is a solution, then $z^4 = a = z_0^4$. As $a \neq 0$, $z_0^4 \neq 0$ so we can divide by $z_0^4$ to obtain 
$$\left(\frac{z}{z_0}\right)^4 = 1.$$ 
Letting $\omega = \frac{z}{z_0}$, we are looking for the solutions to $\omega^4 = 1$ which are the fourth roots of unity $\omega_1, \omega_2, \omega_3, \omega_4$. Therefore $z_1 = \omega_1z_0$, $z_2 = \omega_2z_0$, $z_3 = \omega_3z_0$, and $z_4 = \omega_4z_0$ are solutions to $z^4 = a$. As $z_0 \neq 0$ (because $a \neq 0$) and the fourth roots of unity are distinct, the complex numbers $z_1, z_2, z_3, z_4$ are distinct. As $z^4 = a$ has precisely four solutions in $\mathbb{C}$, we have all the solutions.
The exact same reasoning shows that if $z_0$ is a solution to $z^n = a$, then the solutions are $z_i = \omega_iz_0$, $i = 1, \dots, n$, where $\omega_1, \dots, \omega_n$ are the $n^{\text{th}}$ roots of unity.
A: Multiply by the fourth roots of $1$, i.e $\pm1$, $\pm i$.
