Complex Numbers 1 Question 3
3. Let $z \in C$ be such that $|z|<1$. If $\omega=\frac{5+3 z}{5(1-z)}$, then
(2019 Main, 9 April II)
(a) $4 \operatorname{Im}(\omega)>5$
(b) $5 \operatorname{Re}(\omega)>1$
(c) $5 \operatorname{Im}(\omega)<1$
(d) $5 \operatorname{Re}(\omega)>4$
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Answer:
Correct Answer: 3. (b)
Solution:
- Given complex number
$$ \begin{aligned} & \omega=\frac{5+3 z}{5(1-z)} \\ \Rightarrow & 5 \omega-5 \omega z=5+3 z \\ \Rightarrow \quad & (3+5 \omega) z=5 \omega-5 \\ \Rightarrow \quad & |3+5 \omega||z|=|5 \omega-5| \\ & \quad\left[\text { applying modulus both sides and }\left|z _1 z _2\right|=\left|z _1\right|\left|z _2\right|\right] \\ \therefore & \quad|z|<1 \end{aligned} $$
$$ \Rightarrow \quad \omega+\frac{3}{5}>|\omega-1| $$
Let $\omega=x+i y$, then $x+\frac{3}{5}^{2}+y^{2}>(x-1)^{2}+y^{2}$
$\Rightarrow \quad x^{2}+\frac{9}{25}+\frac{6}{5} x>x^{2}+1-2 x$
$\Rightarrow \frac{16 x}{5}>\frac{16}{25} \Rightarrow x>\frac{1}{5} \Rightarrow 5 x>1$
$\Rightarrow \quad 5 \operatorname{Re}(\omega)>1$
