Ans: C
Note that $x=1$ and $y=1$. Hence, we have
$\begin{array}{rcl}
OA & = & \sqrt{(-1)^2 + 1^2} \\
& = & \sqrt{2}
\end{array}$
and
$\begin{array}{rcl}
\tan \theta & = & \dfrac{-1}{1} \\
\theta & = & 135^\circ
\end{array}$
Therefore, the polar coordinates of $A$ are $(\sqrt{2},135^\circ)$.
