Question

A particle $P$ starts from the point $z_0=1+2i$, where $i=\sqrt{-1}$. It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point $z_1$. From $z_1$ the particle moves $\sqrt{2}$ units in the direction of the vector $\hat{i}+\hat{j}$ and then it moves through an angle $\frac{\pi }{2}$ in anticlockwise direction on a circle with centre at origin, to reach a point $z_2$. The point $z_2$ is given by

• A. $6+7i$
• B. $-7+6i$
• C. $7+6i$
• D. $-6+7i$

Answer 1

Answer: (D)

$-6+7i$

Given $z_o=1+2i$

$⟹z_1=1+2i+5+3i=6+5i$

Let $z_2^{\prime }$ be the point after it moves $\sqrt{2}$ units in the direction $\hat{i}+\hat{j}$. Then,

$⟹z_2^{\prime }=(6+\sqrt{2}\cos {45}^0,5+\sqrt{2}\sin {45}^0)=(7,6)=7+6i$
By rotation about $(0,0),$
$\frac{z_2}{z_2^{\prime }}=e^{\frac{i\pi }{2}}$
$\Rightarrow z_2=z_2^{\prime }\left(e^{\frac{i\pi }{2}}\right)$
$z_2=(7+6i)(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})=(7+6i)\left(i\right)=-6+7i$