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About LinkBacksProve that there are always two square roots of a non zero complex number
How would I do this?
Well that's one way to show it, but I'd compare it to killing a fly with a nuclear bomb!
Every complex number can be written as(polar form). Then
and
are the two square roots of
(
denotes the usual, positive square root of the real number
). I'll let you show that they are distinct when
. That there cannot be any others is a consequence of the fact that polynomials over a field (such as
) cannot have more roots than their degree.
Originally Posted by differentiate
Besides the two answers you already got, you may want to see it in cartesian coordinates:
Now just sove these equations and get two solutions (though it may seem there are 4 you have to choose sign according to...)
Tonio
Yet another way. By DeMoivres' theorem, the kth roots of the complex numberare of the form
for integer j.
Here, k= 2 so the square roots are given by
.
If j is any even number, say j= 2n,so that
If j is odd, say j= 2n+ 1,\pi}{2}}= e^{\frac{4ni\pi}{2}}e^{2i\pi/2})
so
.
The only two square roots areand
.
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