Solution
Given:
V=120(cos4π+isin4π),Z=21+i3
Using Ohm's law V=IZ:
I=ZV
First write Z in polar form:
Z=21+23i
∣Z∣=(21)2+(23)2=1,arg(Z)=tan−1(3)=3π
So:
Z=cos3π+isin3π
Now divide in polar form:
I=1120[cos(4π−3π)+isin(4π−3π)]=120[cos(−12π)+isin(−12π)]
I=120(cos(−12π)+isin(−12π))
Equivalent form:
I=120(cos12π−isin12π)