Answer
The polar coordinate is $\left( \sqrt{2},\frac{7\pi }{4} \right)$.
Work Step by Step
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 $\left( \sqrt{2},\frac{7\pi }{4} \right)$.
