Question

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

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

Answer 1

The correct option is C $5Re\left(\omega \right)>1$

$\left|z\right|<1$

$5\omega (1-z)=5+3z$

$5\omega -5\omega z=5+3z$

$z=\frac{5\omega -5}{3+5\omega }$

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

$|\omega -1|<|3+5\omega |$

$|\omega -1|<5|\omega -(-\frac{3}{5})|$