Imaginaries

[mezarashi]

Half-way through some derivations and I find that I come across an elementary mathematical operation that I embarassingly don't understand. Somehow it seems to have eluded me. The jump from 1 to 2 is:

 

 

[MATH]z = \sqrt{jA^2}[/MATH]

[MATH]z = \sqrt{j}A[/MATH]

 

[MATH]z = \frac{1}{\sqrt{2}} A (1 + j)[/MATH]

 

j indicates the imaginary number. I'm not sure how the square root j was expanded there. Would anybody please explain the idea behind this? Any other generalities to look out for?

[Dave]

Easiest way to do it is by looking at the polar form.

 

[imath]i = \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})[/imath].

 

Now square root: [imath]\sqrt{i} = \left( \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2}) \right)^{\frac{1}{2}}[/imath]

 

Apply de Moivre: [imath]\sqrt{i} = \cos(\frac{\pi}{4}) + i\sin(\frac{\pi}{4}) = \frac{1+i}{\sqrt{2}}[/imath].

[mezarashi]

w00ts alright! Thanks. A bit trickier than I thought... but the text assumes its fundamental =/

[Dave]

I wouldn't say it's a fundamental part of learning complex numbers (I didn't learn about it until the very end of my Further Maths A-level) but it's a nice trick to know. Another way of doing it is to say that [imath]\sqrt{a+bi} \equiv c+di[/imath] for some [imath]c, d \in \C[/imath] and then square both sides, equate co-efficients. That way is a little more tricky (you run into some problems with positive/negative square roots) but works equally as well.