(viii) 32+32i\dfrac{3}{2}+\dfrac{\sqrt{3}}{2}i23+23i r=(32)2+(32)2=3 r=\sqrt{\left(\frac{3}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=\sqrt{3}r=(23)2+(23)2=3 Since x>0x>0x>0 and y>0y>0y>0: θ=tan−1(3232)=tan−1(13)=π6\theta=\tan^{-1}\left(\frac{\frac{\sqrt{3}}{2}}{\frac{3}{2}}\right)=\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}θ=tan−1(2323)=tan−1(31)=6π 32+32i=3(cosπ6+isinπ6)\boxed{\frac{3}{2}+\frac{\sqrt{3}}{2}i=\sqrt{3}\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right)}23+23i=3(cos6π+isin6π)