(vi) 3+i1+3i\dfrac{\sqrt{3}+i}{1+\sqrt{3}i}1+3i3+i 3+i1+3i=(3+i)(1−3i)(1+3i)(1−3i)=23−2i4=32−12i\begin{aligned} \frac{\sqrt{3}+i}{1+\sqrt{3}i} &=\frac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(1+\sqrt{3}i)(1-\sqrt{3}i)}\\ &=\frac{2\sqrt{3}-2i}{4}=\frac{\sqrt{3}}{2}-\frac{1}{2}i \end{aligned}1+3i3+i=(1+3i)(1−3i)(3+i)(1−3i)=423−2i=23−21i So z=32−12iz=\frac{\sqrt{3}}{2}-\frac{1}{2}iz=23−21i. r=(32)2+(−12)2=1 r=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(-\frac{1}{2}\right)^2}=1r=(23)2+(−21)2=1 Since x>0x>0x>0 and y<0y<0y<0 (Quadrant IV): θ=−π6\theta=-\frac{\pi}{6}θ=−6π 3+i1+3i=cos(−π6)+isin(−π6)\boxed{\frac{\sqrt{3}+i}{1+\sqrt{3}i}=\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)}1+3i3+i=cos(−6π)+isin(−6π)