Question

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

• A. ${\sec }^{-1}2$
• B. ${\cot }^{-1}0$
• C. ${\sin }^{-1}\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)$
• D. ${\cos }^{-1}(-\frac{1}{2})$

Answer 1

The correct option is B ${\cot }^{-1}0$

From the graph,
$\arg \left(z_1\right)={\tan }^{-1}\left(5/5\right)={45}^{\circ }$
${\text{cosec}}^{-1}\sqrt{2}={45}^{\circ }$
Since, $z_1$ moves through by an angle ${45}^{\circ }$ in anti-clockwise direction.
The angle made by $z_2$ with axis will be ${45}^{\circ }+{45}^{\circ }={90}^{\circ }$
$\therefore \arg \left(z_2\right)={90}^{\circ }={\cot }^{-1}0$