Question

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

Answer 1

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

Quadrant III

$$\tan \theta=\dfrac{-1}{-\sqrt{3}}=\dfrac{1}{\sqrt{3}}$$ $$\theta=\pi+\tan^{-1}(\dfrac{1}{\sqrt{3}})=\dfrac{7\pi}{6}$$ $$z_1=2(\cos\dfrac{7\pi}{6}+i\sin\dfrac{7\pi}{6})$$