Complex Solutions to Polynomial Equations I have to find another complex solution to the polynomial equation $z^6 - 1 = 0$ given that $\frac{1}{2}−\frac{\sqrt{3}}{2}i$ is a solution. 
I guess the question I have for this is that is there any other way other than expanding the brackets out completely? 
 A: $$x^6-1=(x^3-1)(x^3+1)=(x-1)(x^2+x+1)(x+1)(x^2-x+1)$$
the roots are as follows
$$x_1=1$$
$$x_2=-1$$
roots of $x^2+x+1$ are
$$x_{3,4}=-\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$$
roots of $x^2-x+1$ are
$$x_{5,6}=\frac{1}{2}\pm \frac{\sqrt{3}i}{2}$$
A: There are actually 6 solutions to this equation, called the sixth roots of unity. 
They are as follows: 
$$e^0=1$$
$$e^{i(\pi/3)}=\frac{1}{2}+\frac{\sqrt{3}}{2}$$
$$e^{i(2\pi/3)}=-\frac{1}{2}+\frac{\sqrt{3}}{2}$$
$$e^{i\pi}=-1$$
$$e^{i(4\pi/3)}=-\frac{1}{2}-\frac{\sqrt{3}}{2}$$
$$e^{i(5\pi/3)}=-\frac{1}{2}-\frac{\sqrt{3}}{2}$$
These solutions form a hexagon in the plane. 
How can we derive these solutions?
The first thing to note is that modulus is multiplicative. So, we have the following equation: 
$$\left|z^6\right|=\left|z\right|^6=1$$
From this we conclude that $\left|z\right|=1$. 
Next, note that when you multiply two complex numbers, you add their arguments. So, if $\theta$ is the argument of $z$, we have the following equation: 
$$6\theta=2k\pi \text{ for some integer } k$$
Using these two equations, you can get the six answers mentioned above. 
