Among the statements
(S1): The set $\{ z\in C-\{-i\} :|z| =1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\}$ contains exactly two elements, and
(S2): The set $\{z \in C-\{-1\}:|z| =1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\}$ contains infinitely many elements.
Analyzing Statement S1: Purely Real Condition
We examine statement S1: \( S_1 = \{ z\in \mathbb{C}-\{-i\} :|z| =1 \text{ and } \frac{z-i}{z+i} \text{ is purely real}\} \).
- The condition \( |z|=1 \) means \( z \) lies on the unit circle in the complex plane.
- Let \( w = \frac{z-i}{z+i} \). For \( w \) to be purely real, its imaginary part must be zero, i.e., \( \text{Im}(w) = 0 \).
- Representing \( z \) as \( x+iy \), after algebraic manipulation using \( |z|=1 \), we find \( w = \frac{-ix}{1+y} \).
- The imaginary part \( \frac{-x}{1+y} \) is zero only if \( x=0 \).
- Combining \( x=0 \) with \( |z|=1 \) (\( x^2+y^2=1 \)) yields \( y^2=1 \), meaning \( y=1 \) or \( y=-1 \). Potential solutions are \( z=i \) and \( z=-i \).
- The expression \( \frac{z-i}{z+i} \) is undefined for \( z=-i \). The set definition also explicitly excludes \( z=-i \).
- Thus, the only valid solution is \( z=i \). The set \( S_1 \) contains only one element, \( \{i\} \). Statement S1 is incorrect.
Analyzing Statement S2: Purely Imaginary Condition
We examine statement S2: \( S_2 = \{z \in \mathbb{C}-\{-1\}:|z| =1 \text{ and } \frac{z-1}{z+1} \text{ is purely imaginary}\} \).
- The condition \( |z|=1 \) again means \( z \) lies on the unit circle.
- Let \( w = \frac{z-1}{z+1} \). For \( w \) to be purely imaginary, its real part must be zero, i.e., \( \text{Re}(w) = 0 \).
- Representing \( z \) as \( x+iy \), after algebraic manipulation using \( |z|=1 \), we find \( w = \frac{iy}{1+x} \).
- The real part of \( w \) is \( 0 \). This holds true provided the denominator \( 1+x \neq 0 \).
- The condition \( z \in \mathbb{C}-\{-1\} \) ensures \( x \neq -1 \).
- Therefore, any \( z \) on the unit circle \( |z|=1 \) except \( z=-1 \) satisfies the condition.
- This set is the unit circle excluding the point \( (-1,0) \), which contains infinitely many elements. Statement S2 is correct.
Conclusion
Statement S1 is incorrect because the set contains only one element. Statement S2 is correct because the set contains infinitely many elements. Therefore, only statement S2 is correct.
Similar Questions
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
- The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is
- Let $f$ be a function such that $3f(x) + 2f\left(\frac{m}{19x}\right) = 5x, x \neq 0$, where $m = \sum_{i=1}^{9} (i)^2$. Then $f(5) - f(2)$ is equal to
- The smallest positive integral value of $a$, for which all the roots of $x^4 - ax^2 + 9 = 0$ are real and distinct, is equal to
- $\left(\frac{1}{3} + \frac{4}{7}\right) + \left(\frac{1}{3^2} + \frac{1}{3} \times \frac{4}{7} + \frac{4^2}{7^2}\right) + \left(\frac{1}{3^3} + \frac{1}{3^2} \times \frac{4}{7} + \frac{1}{3} \times \frac{4^2}{7^2} + \frac{4^3}{7^3}\right) + \dots$ upto infinite terms, is equal to
- Let $z = (1 + i)(1 + 2i)(1 + 3i) \dots (1 + ni)$, where $i = \sqrt{-1}$. If $|z|^2 = 44200$, then $n$ is equal to _______
Important Questions from Algebra
- Let $\alpha = \frac{-1 + i\sqrt{3}}{2}$ and $\beta = \frac{-1 - i\sqrt{3}}{2}$, $i = \sqrt{-1}$. If $(7 - 7\alpha + 9\beta)^{20} + (9 + 7\alpha - 7\beta)^{20} + (-7 + 9\alpha + 7\beta)^{20} + (14 + 7\alpha + 7\beta)^{20} = m^{10}$, then $m$ is _________
Let S be the set of the first 11 natural numbers. Then the number of elements in $A = \{B \subseteq S : n(B) \geq 2$ and the product of all elements of B is even is ________.
- The number of $3 \times 2$ matrices A, which can be formed using the elements of the set $\{-2, -1, 0, 1, 2\}$ such that the sum of all the diagonal elements of $A^T A$ is 5, is ______
- The number of numbers greater than 5000, less than 9000 and divisible by 3, that can be formed using the digits 0, 1, 2, 5, 9, if the repetition of the digits is allowed, is ______
- Let $P = [p_{ij}]$ and $Q = [q_{ij}]$ be two square matrices of order 3 such that $q_{ij} = 2^{(i + j - 1)} p_{ij}$ and $\det(Q) = 2^{10}$. Then the value of $\det(\text{adj}(\text{adj } P))$ is:
