Question

Let $z\in C$ be such that $|z|<1$. If $\omega =\frac{5+3z}{5(1-z)}$, then :

• A. $5\text{Re}\left(\omega \right)>4$
• B. $5\text{Re}\left(\omega \right)>1$
• C. $4\text{Im}\left(\omega \right)>5$
• D. $5\text{Im}\left(\omega \right)<1$

Answer 1

The correct option is B $5\text{Re}\left(\omega \right)>1$

$\omega =\frac{5+3z}{5(1-z)}$
$\Rightarrow 5\omega -5\omega z=5+3z$
$\Rightarrow 5\omega -5=5\omega z+3z$
$\Rightarrow z=\frac{5(\omega -1)}{5\omega +3}$

Since, $|z|<1$
$\Rightarrow \left|\frac{5(\omega -1)}{5\omega +3}\right|<1$
$\Rightarrow |5(\omega -1)|<|5\omega +3|$

$\Rightarrow 5|(\omega -1)|<5|\omega +\frac{3}{5}|$
Let $\omega =x+iy$
$\Rightarrow (x-1{)}^2+y^2<{(x+\frac{3}{5})}^2+y^2$
$\Rightarrow x>\frac{1}{5}$
or, $\text{Re}\left(\omega \right)>\frac{1}{5}$