Question

If $|z|=1,|a|\ne 1$, and $x=\frac{z}{(z-a)(1-z\overline{a})}$, then

• A. $x<0$
• B. $x>0$
• C. $0\text{<}x\text{<}1$
• D. $x\ge 1$

Answer 1

Answer: (B)

Given $\left|z\right|=1$ $,$ $\left|a\right|\ne 1$

then, $x=\frac{z}{(z-a)(1-z\overline{a})}=\frac{1}{(1-a)(1-a)}$

let $a=x+iy$ and $\overline{a}=x-iy$

$=\frac{1}{[1-(x+iy)][1-(x-iy)]}$

$=\frac{1}{1-(x-iy)-(x+iy)+(x+iy)(x-iy)}$

$=\frac{1}{1-x+iy-x-iy+x^2-{\left(iy\right)}^2}$

$=\frac{1}{{(x-1)}^2-y^2}$

$\therefore x>0.$

Hence, the answer is $x>0.$