JEE MAIN - Mathematics (2019 - 9th April Evening Slot - No. 5)
Let z $$ \in $$ C be such that |z| < 1.
If $$\omega = {{5 + 3z} \over {5(1 - z)}}$$z, then :
If $$\omega = {{5 + 3z} \over {5(1 - z)}}$$z, then :
4Im( $$\omega$$) > 5
5Im( $$\omega$$) < 1
5Re( $$\omega$$) > 4
5Re( $$\omega$$) > 1
Explanation
$$\omega = {{5 + 3z} \over {5(1 - z)}}$$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 $${\left( { - {3 \over 5}} \right)}$$._9th_April_Evening_Slot_en_5_1.png)
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
$$ \Rightarrow $$ $$5\omega \left( {1 - z} \right) = 5 + 3z$$
$$ \Rightarrow $$ $$5\omega - 5\omega z = 5 + 3z$$
$$ \Rightarrow $$ $$5\omega - 5 = 5\omega z + 3z$$
$$ \Rightarrow $$ $$z\left( {5\omega + 3} \right) = 5\left( {\omega - 1} \right)$$
$$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {3 + 5\omega }}$$
$$ \Rightarrow $$ $$z = {{5\left( {\omega - 1} \right)} \over {5\left( {\omega + {3 \over 5}} \right)}}$$
$$ \Rightarrow $$ $$z = {{\left( {\omega - 1} \right)} \over {\left( {\omega + {3 \over 5}} \right)}}$$
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
Comments (0)
