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 $1 \cap x+i y$
$ \text { or } \quad 1 \leq x \text { and } 0 \leq y $
Given, $\quad \frac{1-z}{1+z} \cap 0 \Rightarrow \frac{1-x-i y}{1+x+i y} \cap 0$
$ \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 \\ \text { and } & -2 y \leq 0 \end{array} $
or $x^{2}+y^{2} \geq 1$ and $y \geq 0$ which is true by Eq. (i).
