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

1. ## 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?

2. Fundamental theorem of algebra - Wikipedia, the free encyclopedia

Kind regards

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

4. 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: $\displaystyle z=x+yi \neq 0 \in \mathbb{C} \Longrightarrow x \neq 0\,\,\,or\,\,\,y \neq 0\,,\,\,so:$

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

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

Tonio

5. Yet another way. By DeMoivres' theorem, the kth roots of the complex number $\displaystyle re^{i\theta}$ are of the form $\displaystyle r^{1/k}e^{\frac{i\theta+ 2ji\pi}{k}}$ for integer j.

Here, k= 2 so the square roots are given by $\displaystyle \sqrt{r}e^{\frac{i\theta+ 2ji\pi}{2}}$$\displaystyle = \sqrt{r}e^{\frac{i\theta}{2}}e^{\frac{2ji\pi}{2}}. If j is any even number, say j= 2n, \displaystyle e^{\frac{2ji\pi}{2}}= e^{4ni\pi/2}= e^{2ni\pi}= 1 so that \displaystyle \sqrt{r}e^{\frac{2ji\pi}{2}}= \sqrt{r} If j is odd, say j= 2n+ 1, \displaystyle e^{\frac{2ji\pi}{2}}= e^{\frac{2i(2n+1)\pi}{2}}= e^{\frac{4ni\pi}{2}}e^{2i\pi/2}$$\displaystyle = e^{2ni\pi}e^{ni\pi}= (1)(-1)= -1$ so $\displaystyle \sqrt{r}e^{\frac{2ji\pi}{2}}= -1\sqrt{r}$.

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