Question

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 $x_1\le x_2$ and $y_1\le y_2$. Then, for all complex numbers $z$ with $1\cap z$, we have $\frac{1-z}{1+z}\cap 0$ then $x^2+y^2\ge 0$ and $y\ge 0$.If true enter 1 else 0.

Answer 1

Let $z=-x+iy\Rightarrow 1\cap z$ gives $1\cap x+iy$
as $1\le x$ and $0\le y$ ...(1)
Given $\frac{1-z}{1+z}\cap 0\Rightarrow \frac{1-x-iy}{1+x+iy}\cap 0\Rightarrow \frac{(1-x-iy)(1+x-iy)}{(1+x+iy)(1+x-iy)}\cap 0+0i$
$\Rightarrow \frac{1-x^2-y^2}{{(1+x)}^2+y^2}-\frac{2iy}{{(1+x)}^2+y^2}\cap 0+0i\Rightarrow x^2+y^2\ge 1$ and $-2y\le 0$
As $x^2+y^2\ge 0$ and $y\ge 0$ which is true by (1)