I need to be able to prove that there are always two square roots of a non-zero complex number, though I'm uncertain as to how I prove this. When I tried doing it algebraically letting $(a+ib)^2 = x+iy$, then getting various simultaneous equations, I end up getting 4 potential roots with all the plus's and minuses.
How do you prove this algebraically?
Is there another method of proving it that is simpler and quicker?
Thanks
On
Using the polar form $z=re^{i\theta}$ makes it easy. You get two roots with the angles different by $\pi$ because $2 \pi i=1$. Your approach should work, but if we don't see your work we can't say what the problem is.
On
Hint: Write your number in the polar form $re^{i\theta}$ instead of the rectangular form $a+ib.$
On
You don't give much context, but if you know (or can assume) the fundamental theorem of algebra that any $n^{th}$ degree polynomial in $\mathbb{C}[X]$ has exactly $n$ roots (counting multiplicity) then the existence of exactly $2$ complex square roots of $c = x+iy$ follows from FTA for $z^2-c=0$.
Expanding your expression:
$$a^2+2abi-b^2=x+iy$$
Comparing real and imaginary parts:
$$a^2-b^2=x\text{ and }2ab=y$$
$$(a^2-b^2)^2+(2ab)^2=x^2+y^2$$
$$a^4-2a^2b^2+b^4+4a^2b^2=x^2+y^2$$
$$a^4+2ab+b^4=x^2+y^2$$
$$(a^2+b^2)^2=x^2+y^2$$
$$a^2+b^2=\sqrt{x^2+y^2}$$
Note: The negative root can be ignored as $a^2+b^2\ge0$
$$a^2=\frac{1}{2}\left(a^2+b^2+a^2-b^2\right)=\frac{1}{2}\left(\sqrt{x^2+y^2}+x\right)$$
$$b^2=\frac{1}{2}\left(a^2+b^2-(a^2-b^2)\right)=\frac{1}{2}\left(\sqrt{x^2+y^2}-x\right)$$
Looking at the right hand side of both these equations the values are both positive so a square root is possible of both:
$$a=\pm\sqrt{\frac{1}{2}\left(\sqrt{x^2+y^2}+x\right)}$$
$$b=\pm\sqrt{\frac{1}{2}\left(\sqrt{x^2+y^2}-x\right)}$$
Now while there are four combinations we know that $2ab=y$ so two of these combinations would be invalid as the product of $ab$ must have the same sign as $y$.