0
$\begingroup$

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

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..

$\endgroup$
1
$\begingroup$

HINT

Note that

$$I^3=j(3\cos45^\circ+3j\sin45^\circ)^2=j(3e^{j\pi/4})^2=9jj=-9$$

thus

$$P=I^3Z=-18-72j \implies r=18\sqrt{17} \quad \theta=\arctan (4) - 180°$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.