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Question: If \(\alpha \) and \(\beta \) are different complex number with \(\left| \alpha \right| = 1\), then ...

If α\alpha and β\beta are different complex number with ∣α∣=1\left| \alpha \right| = 1, then what is ∣α−β1−αβ‾∣\left| {\dfrac{{\alpha - \beta }}{{1 - \alpha \overline \beta }}} \right| equal to?
A. ∣β∣\left| \beta \right|
B. 2
C. 1
D. 0

Explanation

Solution

Hint: Here we will use the conjugate of the given complex numbers to solve.

Complete step-by-step answer:
Multiplying by α‾\overline \alpha on numerator and denominator, we get
∣(α−β)α‾(1−αβ‾)α‾∣=∣(α−β)α‾α‾−αα‾β‾∣\left| {\dfrac{{(\alpha - \beta )\overline \alpha }}{{(1 - \alpha \overline \beta )\overline \alpha }}} \right| = \left| {\dfrac{{(\alpha - \beta )\overline \alpha }}{{\overline \alpha - \alpha \overline \alpha \overline \beta }}} \right|
We know that
z.z‾=∣z∣2 α‾α=∣α∣2=1 ∣(α−β)α‾(α‾−β‾)∣=∣(α−β)(α‾−β‾)∣∣α‾∣  z.\overline z = {\left| z \right|^2} \\\ \overline \alpha \alpha = {\left| \alpha \right|^2} = 1 \\\ \left| {\dfrac{{(\alpha - \beta )\overline \alpha }}{{(\overline \alpha - \overline \beta )}}} \right| = \left| {\dfrac{{(\alpha - \beta )}}{{(\overline \alpha - \overline \beta )}}} \right|\left| {\overline \alpha } \right| \\\
As we know ∣z∣=∣z‾∣\left| z \right| = \left| {\overline z } \right|
Therefore ∣α−β∣=∣α−β‾∣\left| {\alpha - \beta } \right| = \left| {\overline {\alpha - \beta } } \right|
So it gets cancel out,
∣α‾∣=∣α∣=1\left| {\overline \alpha } \right| = \left| \alpha \right| = 1

Note: For modulus type questions in complex numbers, we have to simplify using conjugate and using property of modulus.