Among the statements:(S1) : The set {z in mat | vedclass

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

Question

(C) For $(S1)$: Let $z = x+iy$. Since $|z|=1$,$x^2+y^2=1$. The condition that $\frac{z-i}{z+i}$ is purely real means $\frac{z-i}{z+i} = \overline{\lef

Vedclass · Accepted Answer

only $(S2)$ is correct

Vedclass · Answer

(C) For $(S1)$: Let $z = x+iy$. Since $|z|=1$,$x^2+y^2=1$. The condition that $\frac{z-i}{z+i}$ is purely real means $\frac{z-i}{z+i} = \overline{\left(\frac{z-i}{z+i}ight)} = \frac{\bar{z}+i}{\bar{z}-i}$.$(z-i)(\bar{z}-i) = (z+i)(\bar{z}+i) \Rightarrow |z|^2 - i(z+\bar{z}) - 1 = |z|^2 + i(z+\bar{z}) - 1$.This simplifies to $2i(z+\bar{z}) = 0$,so $z+\bar{z} = 2x = 0$,meaning $x=0$. Since $|z|=1$,$y^2=1$,so $z = \pm i$. However,$z 
eq -i$,so $z=i$ is the only solution. Thus,$(S1)$ is incorrect.For $(S2)$: Let $w = \frac{z-1}{z+1}$. The condition that $w$ is purely imaginary means $w + \bar{w} = 0$.$\frac{z-1}{z+1} + \frac{\bar{z}-1}{\bar{z}+1} = 0 \Rightarrow (z-1)(\bar{z}+1) + (z+1)(\bar{z}-1) = 0$.$|z|^2 + z - \bar{z} - 1 + |z|^2 - z + \bar{z} - 1 = 0 \Rightarrow 2|z|^2 - 2 = 0 \Rightarrow |z|^2 = 1$.Since $|z|=1$ is given,every $z$ on the unit circle (except $z=-1$) satisfies the condition. Thus,$(S2)$ is correct.

Similar Questions

Let the set of all values of $k \in R$ such that the equation $z(\bar{z} + 2 + i) + k(2 + 3i) = 0, z \in C$, has at least one solution, be the interval $[\alpha, \beta]$. Then $9(\alpha + \beta)$ is equal to:

Let $\frac{1 - ix}{1 + ix} = a - ib$ and $a^2 + b^2 = 1$,where $a$ and $b$ are real,then $x = $

$\omega$ is a complex cube root of unity and if $Z$ is a complex number satisfying $|Z-1| \leq 2$ and $|\omega^2 Z-1-\omega|=a$,then the set of possible values of $a$ is

If $a, b \in \mathbb{R}$ and $i=\sqrt{-1}$,then the number of ordered pairs of real numbers $(a, b)$ satisfying the condition $(a+bi)^3 = a-bi$ is

If $(x+iy)^{1/3} = a+ib$ where $x, y, a, b \in R$ and $i = \sqrt{-1}$,then $\frac{x}{a} - \frac{y}{b} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module