## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\left(\sqrt 2, -\frac{\pi}{4}\right)$
Let the polar coordinates be $(r,\theta)$. Then, $r=\sqrt {x^{2}+y^{2}}=\sqrt {(1)^{2}+(-1)^{2}}=\sqrt {2}$ $(1,-1)$ lies in quadrant IV, so to find the angle, use the formula $\theta=\tan^{-1}\frac{y}{x}$ to obtain: $$\theta=\tan^{-1}{\left(\frac{-1}{1}\right)}=-\frac{\pi}{4}$$ A set of polar coordinates for the point $(1,-1)$ is $(\sqrt 2, -\frac{\pi}{4})$.