(iii) 12+32i\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i21+23i r=(12)2+(32)2=1 r=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}=1r=(21)2+(23)2=1 Since x>0x>0x>0 and y>0y>0y>0: θ=tan−1(3)=π3\theta=\tan^{-1}(\sqrt{3})=\frac{\pi}{3}θ=tan−1(3)=3π 12+32i=cosπ3+isinπ3\boxed{\frac{1}{2}+\frac{\sqrt{3}}{2}i=\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}}21+23i=cos3π+isin3π