If $z=\sqrt{2}-i\sqrt{2}$ is rotated through an angle 45° in the anti-clockwise direction about the origin, then the coordinates of its new position are 
[Kerala (Engg.) 2005]

The correct option is E $(\sqrt{2},0$
$z=\sqrt{2}-i\sqrt{2}$
Here, $\theta =ta{n}^{-1}=\left(\frac{-\sqrt{2}}{\sqrt{2}}\right)=ta{n}^{-1}(-1)={135}^o$
Now, rotate z in opposite direction with 45° angle

$\therefore \theta ={180}^o$

$\therefore \theta =ta{n}^0=ta{n}^{-1}\left(\frac{0}{\sqrt{2}}\right)$
$\Rightarrow Hencex=\sqrt{2}andy=0.$

Mathematics