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
Answer
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 $y$.
No comments:
Post a Comment