$$P=I^3Z$$ $$I^3=j(3\cos45^\circ+3j\sin45^\circ)^2$$ $$Z=2+8j$$ $$P=j(3\cos45^\circ+3j\sin45^\circ)^2(2+8j)=j(3\cos(2×45^\circ)+3j\sin(2×45^\circ))×(8.246\angle 75.96)$$
I am trying to convert $j (3 \cos(2×45) + 3j\sin(2×45)) $ to the form of $ z = r\angle \theta$ another form of the polar form of complex numbers.
How do I do that with the $j$ and $3$ in it ?
My attempt is -> $3j (\cos (90) + j\sin(90)) = 3j \angle 90$
This is surely wrong as there shouldn’t be an imaginary part ($j$) in the polar form..