KCET · Maths · Complex Number
If \(\alpha\) and \(\beta\) are different complex numbers with \(|\beta|=1\), then \(\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|\) is equal to
- A \(\frac{1}{2}\)
- B 1
- C \(\frac{1}{3}\)
- D 2
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
Since, \(|\beta|=1 \quad \therefore|\beta|^{2}=\beta \bar{\beta}=1\)
\(\begin{aligned} \therefore \quad\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| &=\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| \\ &=\frac{|\beta-\alpha|}{|\beta||\bar{\beta}-\bar{\alpha}|} \\ &=\frac{|\beta-\alpha|}{1 \cdot|\overline{\beta-\alpha}|} \quad[\because|\overline{\mathrm{z}}|=|\mathrm{z}|] \\ &=1 \end{aligned}\)
\(\begin{aligned} \therefore \quad\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| &=\left|\frac{\beta-\alpha}{\beta \bar{\beta}-\bar{\alpha} \beta}\right| \\ &=\frac{|\beta-\alpha|}{|\beta||\bar{\beta}-\bar{\alpha}|} \\ &=\frac{|\beta-\alpha|}{1 \cdot|\overline{\beta-\alpha}|} \quad[\because|\overline{\mathrm{z}}|=|\mathrm{z}|] \\ &=1 \end{aligned}\)
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