A particle $P$ starts from the point $z_0=1+2 i$, where $i=\sqrt{-1}$. It moves first horizontally away from…
- $6+7 i$
- $-7+6 i$
- $7+6 i$
- $-6+7 i$
Solution

$ z_2^{\prime}=\left(6+\sqrt{2} \times \cos 45^{\circ}, 5+\sqrt{2} \sin 45^{\circ}\right)=(7,6)=7+6 i $ By rotation about $(0,0)$, $ \begin{aligned} & \frac{z_2}{z_2^{\prime}}=e^{i \frac{\pi}{2}} \\ & \Rightarrow \quad z_2=z_2^{\prime}\left(e^{i \frac{\pi}{2}}\right) \\ & \Rightarrow \quad z_2=(7+6 i)\left(\cos \frac{\pi}{-}+i \sin \frac{\pi}{-}\right)=(7+6 i)(i)=-6+7 i \end{aligned} $
Asked in: JEE Advanced 2008 (Paper 2)