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Math Help - Prove that there are always two square roots of a non zero complex number

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    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?
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    MHF Contributor chisigma's Avatar
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    Fundamental theorem of algebra - Wikipedia, the free encyclopedia

    Kind regards

    \chi \sigma
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    MHF Contributor Bruno J.'s Avatar
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    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 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.
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    Quote Originally Posted by differentiate View Post
    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
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    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}}.
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