# Prove that there are always two square roots of a non zero complex number

• Oct 23rd 2009, 10:42 PM
differentiate
Prove that there are always two square roots of a non zero complex number
Prove that there are always two square roots of a non zero complex number

How would I do this?
• Oct 23rd 2009, 10:48 PM
chisigma
Fundamental theorem of algebra - Wikipedia, the free encyclopedia

Kind regards

$\chi$ $\sigma$
• Oct 23rd 2009, 10:59 PM
Bruno J.
Well that's one way to show it, but I'd compare it to killing a fly with a nuclear bomb! (Smirk)

Every complex number can be written as $z=re^{i\theta}$ (polar form). Then $\sqrt{r}e^{i\theta/2}$ and $-\sqrt{r}e^{i\theta/2}$ are the two square roots of $z$ ( $\sqrt r$ denotes the usual, positive square root of the real number $r$). I'll let you show that they are distinct when $z\neq 0$. That there cannot be any others is a consequence of the fact that polynomials over a field (such as $\mathbb{C}$) cannot have more roots than their degree.
• Oct 24th 2009, 05:59 AM
tonio
Quote:

Originally Posted by differentiate
Prove that there are always two square roots of a non zero complex number

How would I do this?

Besides the two answers you already got, you may want to see it in cartesian coordinates: $z=x+yi \neq 0 \in \mathbb{C} \Longrightarrow x \neq 0\,\,\,or\,\,\,y \neq 0\,,\,\,so:$

$(a+bi)^2=x+yi \Longrightarrow a^2-b^2+2abi=x+yi \Longrightarrow \left\{\begin{array}{cc}a^2-b^2=x \\ 2ab=y\end{array}\right.$

Now just sove these equations and get two solutions (though it may seem there are 4 you have to choose sign according to...)

Tonio
• Oct 24th 2009, 07:22 AM
HallsofIvy
Yet another way. By DeMoivres' theorem, the kth roots of the complex number $re^{i\theta}$ are of the form $r^{1/k}e^{\frac{i\theta+ 2ji\pi}{k}}$ for integer j.

Here, k= 2 so the square roots are given by $\sqrt{r}e^{\frac{i\theta+ 2ji\pi}{2}}$ $= \sqrt{r}e^{\frac{i\theta}{2}}e^{\frac{2ji\pi}{2}}$.

If j is any even number, say j= 2n, $e^{\frac{2ji\pi}{2}}= e^{4ni\pi/2}= e^{2ni\pi}= 1$ so that $\sqrt{r}e^{\frac{2ji\pi}{2}}= \sqrt{r}$

If j is odd, say j= 2n+ 1, $e^{\frac{2ji\pi}{2}}= e^{\frac{2i(2n+1)\pi}{2}}= e^{\frac{4ni\pi}{2}}e^{2i\pi/2}$ $= e^{2ni\pi}e^{ni\pi}= (1)(-1)= -1$ so $\sqrt{r}e^{\frac{2ji\pi}{2}}= -1\sqrt{r}$.

The only two square roots are $\sqrt{r}e^{\frac{i\theta}{2}}$ and $-\sqrt{r}e^{\frac{i\theta}{2}}$.