The set of all alpha in R,for which w = frac{ | vedclass

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

Question

(A) Given $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.$w + \bar{w} = \frac{1 + (1 - 8\alpha)z}{1 - z} + \f

Vedclass · Accepted Answer

$\left\{ 0 \right\}$

Vedclass · Answer

(A) Given $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$. Since $w$ is purely imaginary,$w + \bar{w} = 0$.$w + \bar{w} = \frac{1 + (1 - 8\alpha)z}{1 - z} + \frac{1 + (1 - 8\alpha)\bar{z}}{1 - \bar{z}} = 0$$\Rightarrow (1 + (1 - 8\alpha)z)(1 - \bar{z}) + (1 + (1 - 8\alpha)\bar{z})(1 - z) = 0$$\Rightarrow 1 - \bar{z} + (1 - 8\alpha)z - (1 - 8\alpha)z\bar{z} + 1 - z + (1 - 8\alpha)\bar{z} - (1 - 8\alpha)z\bar{z} = 0$Since $|z| = 1$,we have $z\bar{z} = 1$.$\Rightarrow 2 - (z + \bar{z}) + (1 - 8\alpha)(\bar{z} + z) - 2(1 - 8\alpha) = 0$$\Rightarrow 2 - (z + \bar{z}) + (1 - 8\alpha)(z + \bar{z}) - 2 + 16\alpha = 0$$\Rightarrow (z + \bar{z})(1 - 8\alpha - 1) + 16\alpha = 0$$\Rightarrow (z + \bar{z})(-8\alpha) + 16\alpha = 0$$\Rightarrow -8\alpha(z + \bar{z} - 2) = 0$Since $	ext{Re}(z) 
eq 1$,$z + \bar{z} 
eq 2$,so $z + \bar{z} - 2 
eq 0$.Therefore,$-8\alpha = 0$,which implies $\alpha = 0$.

The set of all $\alpha \in R$,for which $w = \frac{1 + (1 - 8\alpha)z}{1 - z}$ is a purely imaginary number,for all $z \in C$ satisfying $|z| = 1$ and $\text{Re}(z) \neq 1$,is

Similar Questions

If $a = \operatorname{Im}\left(\frac{1+z^2}{2iz}\right)$ and $z$ is any non-zero complex number such that $|z|=1$,then $a=$

If $C$ is a point on the straight line joining the points $A(-2+i)$ and $B(3-4i)$ in the Argand plane and $\frac{AC}{CB}=\frac{1}{2}$,then the argument of $C$ is

Let $S = \{z \in \mathbb{C} : |z-3| \leq 1 \text{ and } z(4+3i) + \bar{z}(4-3i) \leq 24\}$. If $\alpha + i\beta$ is the point in $S$ which is closest to $4i$,then $25(\alpha + \beta)$ is equal to

What are the polar coordinates of the point $(-\sqrt{3}, 1)$?

If $z_1, z_2, z_3 \in \mathbb{C}$ such that $|z_1| = |z_2| = |z_3| = 2$,then the greatest value of the expression $|z_1 - z_2||z_2 - z_3| + |z_2 - z_3||z_3 - z_1| + |z_3 - z_1||z_1 - z_2|$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module