## Precalculus (6th Edition) Blitzer

The polar coordinate is $\left( \sqrt{2},\frac{7\pi }{4} \right)$.
Consider the given point $\left( 1,-1 \right)$ ; So, \begin{align} & r=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}} \\ & =\sqrt{1+1} \\ & =\sqrt{2} \end{align} Also, \begin{align} & \tan \theta =\frac{-1}{1} \\ & =-1 \end{align} As $\tan \frac{\pi }{4}=1$ and $\theta$ lies in the fourth quadrant. So, \begin{align} & \theta =2\pi -\frac{\pi }{4} \\ & =\frac{7\pi }{4} \end{align} So, the polar coordinate is $\left( \sqrt{2},\frac{7\pi }{4} \right)$.