Let point P = alpha + ibeta,where alpha, beta | vedclass

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

Question

(D) Let the initial point be $P = \alpha + i\beta$.$(I)$ Reflection of $z = x + iy$ about $	ext{amp}(z) = \frac{\pi}{4}$ is $z' = i\bar{z} = i(x - iy

Vedclass · Accepted Answer

$\alpha = \sqrt{3} + 1$

Vedclass · Answer

(D) Let the initial point be $P = \alpha + i\beta$.$(I)$ Reflection of $z = x + iy$ about $	ext{amp}(z) = \frac{\pi}{4}$ is $z' = i\bar{z} = i(x - iy) = y + ix$. Thus,$P_1 = \beta + i\alpha$.$(II)$ Shifting $P_1$ by $\beta$ units along the positive real axis gives $P_2 = (\beta + \beta) + i\alpha = 2\beta + i\alpha$.$(III)$ Rotating $P_2$ by $\frac{\pi}{4}$ counter-clockwise about the origin gives $Q = P_2 \cdot e^{i\pi/4} = (2\beta + i\alpha) \left( \frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}} ight)$.Given $Q = -\sqrt{2} + i\sqrt{6}$,we have:$-\sqrt{2} + i\sqrt{6} = \frac{1}{\sqrt{2}} (2\beta - \alpha) + i \frac{1}{\sqrt{2}} (2\beta + \alpha)$.Equating real and imaginary parts:$2\beta - \alpha = -2$  $(1)$$2\beta + \alpha = 2\sqrt{3}$ $(2)$Adding $(1)$ and $(2)$: $4\beta = 2\sqrt{3} - 2 \implies \beta = \frac{\sqrt{3} - 1}{2}$.Subtracting $(1)$ from $(2)$: $2\alpha = 2\sqrt{3} + 2 \implies \alpha = \sqrt{3} + 1$.Thus,the correct option is $D$.

Similar Questions

If $z = x + iy$ and $|z - 2 + i| = |z - 3 - i|$,then the locus of $z$ is:

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

The set of all real values of $c$ for which the equation $z \bar{z} + (4 - 3i) \bar{z} + (4 + 3i) z + c = 0$ represents a circle is

The minimum value of the expression $|z|+|z-1|+|z-1-i|+|z-i|$,where $z$ is a complex number and $i=\sqrt{-1}$,is

Let $z_{1}$ and $z_{2}$ be two complex numbers such that $\arg(z_{1}-z_{2})=\frac{\pi}{4}$ and $z_{1}, z_{2}$ satisfy the equation $|z-3|=\operatorname{Re}(z)$. Then the imaginary part of $z_{1}+z_{2}$ is equal to ..... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module