Let z C be such that |z|

Question

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

Accepted Answer

$$\omega = {{5 + 3z} \over {5(1 - z)}}$$z $$ \Rightarrow $$ $$5\omega \left( {1 - z} \right) = 5 + 3z$$ $$ \Rightarrow $$ $$5\omega - 5\omega z = 5 + 3z$$ $$ \Rightarrow $$ $$5\omega - 5 = 5\omega z + 3z$$ $$ \Rightarrow $$ $$z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$$ $$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$$ $$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {5\left( {\omega + {3 \over 5}} \right)}}$$ $$ \Rightarrow $$ $$z = {{\left( {\omega - 1} \right)} \over {\left( {\omega + {3 \over 5}} \right)}}$$ Given |z| < 1 $$ \Rightarrow $$ $$\left| {{{\omega - 1} \over {\omega + {3 \over 5}}}} \right|$$ < 1 $$ \Rightarrow $$ $$\left| {\omega - 1} \right| < \left| {\omega + {3 \over 5}} \right|$$ $$ \Rightarrow $$ $$\left| {\omega - 1} \right| < \left| {\omega - \left( { - {3 \over 5}} \right)} \right|$$ It says distance of $$\omega $$ from point 1 is less than distance from point $${\left( { - {3 \over 5}} \right)}$$. x coordinate of perpendicular bisector of points $${\left( { - {3 \over 5},0} \right)}$$ and (1, 0), x = $${{1 - {3 \over 5}} \over 2}$$ = $${1 \over 5}$$ As $$\omega $$ is closer to point (1, 0) so $$\omega $$ should present in the right side of perpendicular bisector. $$ \therefore $$ $${\mathop{\rm Re}\nolimits} (\omega ) > {1 \over 5}$$ $$ \Rightarrow $$ 5Re( $$\omega$$) > 1 Other Method : $$\omega = {{5 + 3z} \over {5(1 - z)}}$$z $$ \Rightarrow $$ $$5\omega \left( {1 - z} \right) = 5 + 3z$$ $$ \Rightarrow $$ $$5\omega - 5\omega z = 5 + 3z$$ $$ \Rightarrow $$ $$5\omega - 5 = 5\omega z + 3z$$ $$ \Rightarrow $$ $$z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$$ $$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$$ Given |z| < 1 $$ \Rightarrow $$ $$\left| {{{5\left( {\omega - 1} \right)} \over {5\omega + 3}}} \right|$$ < 1 $$ \Rightarrow $$ $$5\left| {\omega - 1} \right| < \left| {5\omega + 3} \right|$$ $$ \Rightarrow $$ $$25\left( {{\omega ^2} - 2\omega + 1} \right) < 25{\omega ^2} + 30\omega + 9$$ $$ \Rightarrow $$ $$25\left( {\omega \overline \omega - \omega - \overline \omega + 1} \right)$$ < $$25\omega \overline \omega + 15\omega + 15\overline \omega + 9$$ (Using $${\left| z \right|^2} = z\overline z $$) $$ \Rightarrow $$ $$16 < 40\omega + 40\overline \omega $$ $$ \Rightarrow $$ $$\omega + \overline \omega > {2 \over 5}$$ $$ \Rightarrow $$ $$2{\mathop{\rm Re}\nolimits} \left( \omega \right) > {2 \over 5}$$ $$ \Rightarrow $$ 5Re( $$\omega$$) > 1

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