# Phasor math for balanced 3 phase circuit

Are there any mathematicians in here? I am currently reading my ECE text's chapter on AC machines and i referred to some material in the chapter that first discussed AC circuits.

In the photo below of the math, I am unsure of how on the right hand side, the subtraction of two phasors became a multiplication of 2 phasors... Specifically, following the red arrow, how did the term furthest to the right go from the form it had I*<150-theta to (1 - 1*<120) ??

I do not recall seeing this anywhere else in the chapter or on the review of complex arithmetic in the appendix and it is sort of bugging me. I am not seeing how they got from that line to the next line in the solution process.

• you should ask this in math.stackexchange.com site, not here.That forum is totally dedicated to mathematical problems. Jun 21, 2016 at 22:40

First of all, multiplication of two complex number is multiplication of there magnitude and addition of there angle. like this: $$I_{\Delta}\angle150 ^{o} =I_{\Delta}\angle120 ^{o}\times 1\angle30 ^{o}$$ Now: $$(I_{\Delta}\angle30^{o}-\theta ^{o})-(I_{\Delta}\angle150 ^{o}-\theta ^{o}) =(I_{\Delta}\angle30^{o}-\theta ^{o})-(1\angle120 ^{o} \times (I_{\Delta}\angle30 ^{o}-\theta ^{o})) =(I_{\Delta}\angle30^{o}-\theta ^{o})\times(1-1\angle120 ^{o})$$