Answer to Question #350492 in Calculus for Nomsa

Question

Convert (√3,-1)into polar coordinates (r,0) so that r≥0 and 0≤0<2π

Answer 1

$$r^2=(\sqrt{3})^2+(-1)^2=4$$ $$r\ge0=>r=\sqrt{4}=2$$

Quadrant IV

$$\tan \theta=\dfrac{-1}{\sqrt{3}}=-\dfrac{1}{\sqrt{3}}$$ $$\theta=2\pi+\tan^{-1}(-\dfrac{1}{\sqrt{3}})=\dfrac{11\pi}{6}$$ $$z=2(\cos\dfrac{11\pi}{6}+i\sin\dfrac{11\pi}{6})$$

$(2, \dfrac{11\pi}{6})$