Question

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

• A. $5Im(\omega )<1$
• B. $4Im(\omega )>5$
• C. $5Re(\omega )>1$
• D. $5Re(\omega )>4$

Answer 1

Answer: (C)

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