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