# How to convert $j(3\cos 45 + 3j\sin 45)$ form to $z = r\angle \theta$

$$P=I^3Z$$ $$I^3=j(3\cos45^\circ+3j\sin45^\circ)^2$$ $$Z=2+8j$$ $$P=j(3\cos45^\circ+3j\sin45^\circ)^2(2+8j)=j(3\cos(2×45^\circ)+3j\sin(2×45^\circ))×(8.246\angle 75.96)$$

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..

$$I^3=j(3\cos45^\circ+3j\sin45^\circ)^2=j(3e^{j\pi/4})^2=9jj=-9$$
$$P=I^3Z=-18-72j \implies r=18\sqrt{17} \quad \theta=\arctan (4) - 180°$$