Exercise [04.03]

JoshSmith · 25 Nov 2009, 08:10

Hopefully this helps some people (like me) rusty with their algebra.

Show $a+ib=\left(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\right)^2$ .

This requires a fair amount of algebraic expansion:

$a+ib=\Bigg(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\Bigg)^2$
$=\Bigg(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\Bigg)\cdot\Bigg(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\Bigg)$
$=\frac{1}{2}(a+\sqrt{a^2+b^2})+2\left(i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2}})\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}\right)+i^2\frac{1}{2}(-a+\sqrt{a^2+b2})$
$=\frac{1}{2}(a+\sqrt{a^2+b^2})+2i\left(\sqrt{(-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})\right)-\frac{1}{2}(-a+\sqrt{a^2+b2})$
$=\left(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2}+\frac{a}{2}-\frac{\sqrt{a^2+b2}}{2}\right)+2i\left(\sqrt{(-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})\right)$
$=a+2i\left(\sqrt{(-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})\right)$
$=a+2i\sqrt{-\frac{a^2}{4}+\frac{a^2+b^2}{4}}$
$=a+2i\sqrt{\frac{b^2}{4}}$
$=a+2i\frac{b}{2}$
=a+ib~,

as we intended to show.

I've intentionally left out some intermediary steps. If you can't work through these intermediaries, I highly suggest doing some more work in algebra. I.M. Gelfand's text has served me well.