Q44 / 90 · KVPY 2016 (SB/SX) · Mathematics
Let a be a fixed non-zero complex number with $|\mathrm{a}| < 1$ and $\mathrm{w}=\left(\frac{\mathrm{z}-\mathrm{a}}{1-\overline{\mathrm{a}} \mathrm{z}}\right)$. Where $\mathrm{z}$ is a complex number. Then
- Athere exists a complex number $\mathrm{z}$ with $|\mathrm{z}| < 1$ such that $|\mathrm{w}|>1$
- B$|\mathrm{w}|>1$ for all $\mathrm{z}$ such that $|\mathrm{z}| < 1$
- Correct answerC$|\mathrm{w}| < 1$ for all $\mathrm{z}$ such that $|\mathrm{z}| < 1$
- Dthere exists $\mathrm{z}$ such with $|\mathrm{z}| < 1$ and $|\mathrm{w}|=1$
Solution
$|\mathrm{a}| < 1 \& \mathrm{w}=\left(\frac{\mathrm{z}-\mathrm{a}}{1-\overline{\mathrm{a}} \mathrm{z}}\right) \Rightarrow \mathrm{w}-\overline{\mathrm{a}} \mathrm{zw}=\mathrm{z}-\mathrm{a}$
$\Rightarrow \mathrm{w}+\mathrm{a}=\mathrm{z}(1+\overline{\mathrm{a}} \mathrm{w})$
$\mathrm{z}=\frac{\mathrm{w}+\mathrm{a}}{1+\overline{\mathrm{a}} \mathrm{w}}$
Given $|\mathrm{z}| < 1$
$\left|\frac{\mathrm{w}+\mathrm{a}}{1+\overline{\mathrm{a}} \mathrm{w}}\right| < 1 \Rightarrow|\mathrm{w}+\mathrm{a}|^{2} < |1+\overline{\mathrm{a}} \mathrm{w}|^{2}$
$\Rightarrow(\mathrm{w}+\mathrm{a})(\overline{\mathrm{w}}+\overline{\mathrm{a}}) < (1+\overline{\mathrm{a}} \mathrm{w})(1+\mathrm{a} \overline{\mathrm{w}})$
$\Rightarrow \mathrm{w} \overline{\mathrm{w}}+\mathrm{w} \overline{\mathrm{a}}+\mathrm{a} \overline{\mathrm{w}}+\mathrm{a} \overline{\mathrm{a}} < 1+\overline{\mathrm{a}} \mathrm{w}+\mathrm{a} \overline{\mathrm{w}}+\mathrm{a} \overline{\mathrm{a}} \mathrm{w} \overline{\mathrm{w}}$
$\Rightarrow \mathrm{a} \overline{\mathrm{a}} \cdot \mathrm{w} \overline{\mathrm{w}}-\mathrm{w} \overline{\mathrm{w}}-\mathrm{a} \overline{\mathrm{a}}+1>0$
$\Rightarrow|\mathrm{a}|^{2}|\mathrm{w}|^{2}-|\mathrm{w}|^{2}|\mathrm{a}|^{2}+1>0$
$\Rightarrow\left(|\mathrm{a}|^{2}-1\right)\left(|\mathrm{w}|^{2}-1\right)>0$
Given $|\mathrm{a}| < 1 \quad|\mathrm{w}|^{2}-1 < 0$
$|\mathrm{w}| < 1 \&|\mathrm{z}| < 1$
Hence, the correct answer is option: 'C'.