(iv) −52−532i-\dfrac{5}{2}-\dfrac{5\sqrt{3}}{2}i−25−253i r=(−52)2+(−532)2=5 r=\sqrt{\left(-\frac{5}{2}\right)^2+\left(-\frac{5\sqrt{3}}{2}\right)^2}=5r=(−25)2+(−253)2=5 Both x<0x<0x<0 and y<0y<0y<0, so the point lies in Quadrant III. tan−1(yx)=tan−1(3)=π3⇒θ=π3+π=4π3\tan^{-1}\left(\frac{y}{x}\right)=\tan^{-1}(\sqrt{3})=\frac{\pi}{3} \Rightarrow \theta=\frac{\pi}{3}+\pi=\frac{4\pi}{3}tan−1(xy)=tan−1(3)=3π⇒θ=3π+π=34π −52−532i=5(cos4π3+isin4π3)\boxed{-\frac{5}{2}-\frac{5\sqrt{3}}{2}i=5\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)}−25−253i=5(cos34π+isin34π)