If $\theta$ is the angle of rotation, then the co-ordinates in the new system are $x^{\prime}=x \cos \theta+y \sin \theta$, $y^{\prime}=y \cos \theta-x \sin \theta$
Given that $x^{\prime}=\sqrt{2}, y^{\prime}=4$
Thus, $x \cos \theta+y \sin \theta=\sqrt{2}$
$$
y \cos \theta-x \sin \theta=4
$$
Also, $\theta=\frac{\pi}{4} \Rightarrow x \cos \frac{\pi}{4}+y \sin \frac{\pi}{4}=\sqrt{2}$
and $y \cos \frac{\pi}{4}-x \sin \frac{\pi}{4}=4$
$$
\begin{array}{ll}
\Rightarrow & x+y=2 \\
\text { and } & y-x=4 \sqrt{2}
\end{array}
$$
On adding Eqs. (i) and (ii), we get $2 y=2+4 \sqrt{2}$ $\Rightarrow \quad y=1+2 \sqrt{2}$
On subtracting (i) and (ii), we get
$$
\begin{array}{rlrl}
& & 2 x & =2-4 \sqrt{2} \\
& x & =1-2 \sqrt{2}
\end{array}
$$
$$
\Rightarrow \quad x=1-2 \sqrt{2}
$$
Thus, the co-ordinates of $(\sqrt{2}, 4)$ in the old system $(1-2 \sqrt{2}, 1+2 \sqrt{2})$