I am taking a power engineering class and I am not sure about taking the squared magnitude of the complex numbers. I figured out most of numbers, except -54.24Vrfl. Please click on the picture: enter image description here. Can someone tell me how did I get it? Thanks a bunch!
$441.7\angle\delta=(.9313\text{Vrfl}-30.04)+j(.0034\text{Vrfl}+251.97).$.
Taking the squared magnitudes of both sides,
$441.7^2=.8673\text{Vrfl}^2-54.24\text{Vrfl}+64,391.$
Btw, Vrfl is voltage.
$$|(.9313\text{Vrfl}-30.04)+j(.0034\text{Vrfl}+251.97)|^2=\\(.9313\text{Vrfl}-30.04)^2+(.0034\text{Vrfl}+251.97)^2=\\ (0.8673\text{Vrfl}^2-55.953\text{Vrfl}+902.4)+(0.00001\text{Vrfl}^2+1.713\text{Vrfl}+63488.88)=\\ 0.8673\text{Vrfl}^2-54.24\text{Vrfl}+64391.28 $$