JEE Main 2019 — Mathematics Algebra

Given $|z|<1$∣z∣<1 and $\omega =\frac{5+3z}{5(1-z)}$ω=5(1−z)5+3z​

$$\Rightarrow 5\omega (1-z)=5+3z$$⇒5ω(1−z)=5+3z

$$\Rightarrow 5\omega -5\omega z=5+3z$$⇒5ω−5ωz=5+3z

$$\Rightarrow z=\frac{5\omega -5}{3+5\omega }$$⇒z=3+5ω5ω−5​

Using the given condition $|z|=5|\frac{\omega -1}{3+5\omega }|<1$∣z∣=5∣3+5ωω−1​∣<1

$$\Rightarrow 5|\omega -1|<|3+5\omega |$$⇒5∣ω−1∣<∣3+5ω∣

$$\Rightarrow 5|\omega -1|<5|\omega +\frac{3}{5}|$$⇒5∣ω−1∣<5∣ω+53​∣

$$\Rightarrow |\omega -1|<|\omega -(-\frac{3}{5})|$$⇒∣ω−1∣<∣ω−(−53​)∣

We know that the locus of a complex number ${z}_{1}$z1​ satisfying $|{z}_{1}-a|=|{z}_{2}-b|$∣z1​−a∣=∣z2​−b∣ is the perpendicular bisector of the line segment joining the points $a$a and $b.$b.

Hence, the locus of the point $\omega$ω satisfying $|\omega -1|=|\omega -(-\frac{3}{5})|$∣ω−1∣=∣ω−(−53​)∣ is the line $x=\frac{1-\frac{3}{5}}{2}=\frac{1}{5}$x=21−53​​=51​

Hence, $Re(\omega )>\frac{1}{5}$Re(ω)>51​

$$\Rightarrow 5Re(\omega )>1.$$⇒5Re(ω)>1.