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Square roots of complex numbers

by dalcde
Tags: complex, numbers, roots, square
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dalcde
#1
Sep16-11, 05:46 AM
P: 166
In general, how many square roots does a complex number have?
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CompuChip
#2
Sep16-11, 07:01 AM
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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 [itex]z = r e^{i\phi}[/itex] then
[tex]u_1 = \sqrt{r} e^{i\phi / 2} = \sqrt{r} \left( \cos \frac{\phi}{2} + i \sin \frac{\phi}{2} \right)[/tex]
and
[tex]u_2 = -u_1[/tex]
both square to z.
phyzguy
#3
Sep16-11, 07:03 AM
P: 2,179
Two - try looking up the Fundamental Theorem of Algebra.

HallsofIvy
#4
Sep16-11, 07:51 AM
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Thanks
PF Gold
P: 39,363
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 [itex]z= re^{i\theta}[/itex] with r> 0. Then the nth roots of z are given by [itex]r^{1/n}e^{i(\theta+ 2k\pi)/n}[/itex] where [itex]r^{1/n}[/itex] 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 [itex]0\le 2k\pi/n< 2\pi[/itex] but when k= n, [itex]2k\pi/n= 2n\pi/n= 2\pi[/itex] and [itex]e^{i(\theta+ 2\pi)}= e^{i\theta}[/itex].


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