Square roots of complex numbers

by dalcde
Tags: complex, numbers, roots, square
 P: 166 In general, how many square roots does a complex number have?
 HW Helper Sci Advisor P: 4,281 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.
 P: 1,957 Two - try looking up the Fundamental Theorem of Algebra.
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Square roots of complex numbers

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 $e^{i(\theta+ 2\pi)}= e^{i\theta}$.

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