The point represented by 2 + i in the Argand | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The initial position is $z_0 = 2 + i$,which corresponds to the point $(2, 1)$ in the Argand plane.Moving $1 \, 	ext{unit}$ eastwards changes the

Vedclass · Accepted Answer

$1+i$

Vedclass · Answer

(A) The initial position is $z_0 = 2 + i$,which corresponds to the point $(2, 1)$ in the Argand plane.Moving $1 \, 	ext{unit}$ eastwards changes the position to $(2+1, 1) = (3, 1)$.Moving $2 \, 	ext{units}$ northwards changes the position to $(3, 1+2) = (3, 3)$.Finally,moving $2\sqrt{2} \, 	ext{units}$ in the south-westwards direction means moving $2 \, 	ext{units}$ west and $2 \, 	ext{units}$ south (since the displacement vector is $(-2\sqrt{2} \cos(45^{\circ}), -2\sqrt{2} \sin(45^{\circ})) = (-2, -2)$).The final position is $(3-2, 3-2) = (1, 1)$.This corresponds to the complex number $1 + i$.

The point represented by $2 + i$ in the Argand plane moves $1 \, \text{unit}$ eastwards,then $2 \, \text{units}$ northwards and finally from there $2\sqrt{2} \, \text{units}$ in the south-westwards direction. Then its new position in the Argand plane is at the point represented by

Similar Questions

If $P(x, y)$ denotes $z = x + iy$ where $x, y \in \mathbb{R}$ and $i = \sqrt{-1}$ in the Argand plane,and $\left|\frac{z-1}{z+2i}\right| = 1$,then the locus of $P$ is

Let $z$ be a complex number. Then the angle between vectors $z$ and $-iz$ is

$\omega$ is a complex cube root of unity and $Z$ is a complex number satisfying $|Z-1| \leq 2$. The possible values of $r$ such that $|Z-1| \leq 2$ and $|\omega Z - 1 - \omega^2| = r$ have no common solution are

If $w = \frac{z}{z - \frac{1}{3}i}$ and $|w| = 1$,where $i = \sqrt{-1}$,then $z$ lies on

If $z_1$ and $z_2$ are two complex numbers satisfying the equation $\left|\frac{z_1+z_2}{z_1-z_2}\right|=1$, then $\frac{z_1}{z_2}$ may be

Vedclass Test Series

Exam Paper Generator

Online Exam Module