Solution Given: Z=3−6i,V=90+30iZ=3-6i,\quad V=90+30iZ=3−6i,V=90+30i Rectangular form I=VZ=90+30i3−6i⋅3+6i3+6iI=\frac{V}{Z}=\frac{90+30i}{3-6i}\cdot\frac{3+6i}{3+6i}I=ZV=3−6i90+30i⋅3+6i3+6i Denominator: (3−6i)(3+6i)=32+62=45(3-6i)(3+6i)=3^2+6^2=45(3−6i)(3+6i)=32+62=45 Numerator: (90+30i)(3+6i)=270+540i+90i+180i2=270+630i−180=90+630i\begin{aligned} (90+30i)(3+6i) &=270+540i+90i+180i^2\\ &=270+630i-180\\ &=90+630i \end{aligned}(90+30i)(3+6i)=270+540i+90i+180i2=270+630i−180=90+630i So: I=90+630i45=2+14iI=\frac{90+630i}{45}=2+14iI=4590+630i=2+14i I=2+14i amperes\boxed{I=2+14i\ \text{amperes}}I=2+14i amperes Polar form ∣I∣=22+142=200=102|I|=\sqrt{2^2+14^2}=\sqrt{200}=10\sqrt{2}∣I∣=22+142=200=102 Angle (Quadrant I): θ=tan−1(142)=tan−1(7)\theta=\tan^{-1}\left(\frac{14}{2}\right)=\tan^{-1}(7)θ=tan−1(214)=tan−1(7) I=102(cos(tan−17)+isin(tan−17))\boxed{I=10\sqrt{2}\left(\cos(\tan^{-1}7)+i\sin(\tan^{-1}7)\right)}I=102(cos(tan−17)+isin(tan−17))