Square roots of complex numbers

In general, how many square roots does a complex number have?
 PhysOrg.com mathematics news on PhysOrg.com >> Pendulum swings back on 350-year-old mathematical mystery>> Bayesian statistics theorem holds its own - but use with caution>> Math technique de-clutters cancer-cell data, revealing tumor evolution, treatment leads
 Blog Entries: 6 Recognitions: Homework Help Science Advisor 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.
 Two - try looking up the Fundamental Theorem of Algebra.

Recognitions:
Gold Member
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 $e^{i(\theta+ 2\pi)}= e^{i\theta}$.