Question

If $|z|=5$ and $w=\frac{z-5}{z+5}$ , then Re $(w)$ is equal to

• A. 0
• B. 1/25
• C. 25
• D. 1
• E. -1

Answer 1

Answer: (A)

Let $z=x+iy$
$\therefore |z|=\sqrt{x^2+y^2}$
$\Rightarrow \sqrt{x^2+y^2}=5$
$\Rightarrow x^2+y^2=25...(i)$
Now, $w=\frac{z-5}{z+5}=\frac{x+iy-5}{x+iy+5}$
$=\frac{(x-b)+iy}{(x+5)+iy}$
$=\frac{(x-5)+iy}{(x+5)+iy}\times \frac{(x+5)-iy}{(x+5)-iy}$
$=\frac{x^2-25+iy(x+5)-y(x-5)i+y^2}{(x+5{)}^2+y^2}$
$=\frac{\left(x^2+y^2-25\right)+i[xy+5y-xy+5y]}{(x+5{)}^2+y^2}$
$=\frac{0+10yi}{(x+5{)}^2+y^2}[\because$ from Eq. (i) $]$
$=\frac{10y}{(x+5{)}^2+y^2}i$
$\therefore$ Re$(w)=0$