SATHEE: Complex Numbers 2 Question 36

Complex Numbers 2 Question 36

37.

For complex numbers $z_{1} = x_{1} + i y_{1}$ and $z_{2} = x_{2} + i y_{2}$ , we write $z_{1} \cap z_{2}$ , if $x_{1} \leq x_{2}$ and $y_{1} \leq y_{2}$ . Then, for all complex numbers $z$ with $1 \cap z$ , we have $\frac{1 - z}{1 + z} \cap 0$ .

$(1981, 2 M)$

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Answer:

Correct Answer: 37. True

Solution:

Let $z = x + i y \Rightarrow 1 \cap z$ gives $1 \cap x + i y$

$or 1 \leq x and 0 \leq y$

Given, $\frac{1 - z}{1 + z} \cap 0 \Rightarrow \frac{1 - x - i y}{1 + x + i y} \cap 0$

$\begin{array}{ll} \Rightarrow & \frac{(1 - x - i y) (1 + x - i y)}{(1 + x + i y) (1 + x - i y)} \cap 0 + 0 i \\ \Rightarrow & \frac{1 - x^{2} - y^{2}}{(1 + x)^{2} + y^{2}} - \frac{2 i y}{(1 + x)^{2} + y^{2}} \cap 0 + 0 i \\ \Rightarrow & x^{2} + y^{2} \geq 1 \\ and & - 2 y \leq 0 \end{array}$

or $x^{2} + y^{2} \geq 1$ and $y \geq 0$ which is true by Eq. (i).

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Complex Numbers 2 Question 36

37.

Answer:

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