# How many square roots does a complex number have?

1. Sep 16, 2011

### dalcde

In general, how many square roots does a complex number have?

2. Sep 16, 2011

### CompuChip

A square root u of z is the solution to the polynomial equation
u2 = z
(where z must be considered as a fixed number).
It is a general theorem that this second-degree polynomial has two complex roots.

In fact, you can write them down explicitly: if $z = r e^{i\phi}$ then
$$u_1 = \sqrt{r} e^{i\phi / 2} = \sqrt{r} \left( \cos \frac{\phi}{2} + i \sin \frac{\phi}{2} \right)$$
and
$$u_2 = -u_1$$
both square to z.

3. Sep 16, 2011

### phyzguy

Two - try looking up the Fundamental Theorem of Algebra.

4. Sep 16, 2011

### HallsofIvy

In fact, it is easy to show that any non-zero complex number has precisely n distinct nth roots:

Let $z= re^{i\theta}$ with r> 0. Then the nth roots of z are given by $r^{1/n}e^{i(\theta+ 2k\pi)/n}$ where $r^{1/n}$ is the positive real nth root of the positive real number r and k is a non-negative integer.

For k= 0 to n-1, those are distinct because $0\le 2k\pi/n< 2\pi$ but when k= n, $2k\pi/n= 2n\pi/n= 2\pi$ and $e^{i(\theta+ 2\pi)}= e^{i\theta}$.