Question

If α and β are different complex numbers with $\left| \beta \right| = 1$, then find $\left| {\dfrac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$.

Answer 1

Complex number can be written as $z = a + ib$. Where a, b are the real numbers.
Modulus of a complex number can be calculated as $\left| z \right| = \sqrt {{a^2} + {b^2}}$
Conjugate of a complex number can be calculated as $\overline z = a - ib$

Complete step-by-step answer:
it is given that, $\left| \beta \right| = 1$
We have to find the values of $\left| {\dfrac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$. Where α and β are complex numbers.
Step1
let $\alpha = a + ib$; a, b are the real numbers.
And $\beta = p + iq$ ; p, q are the real numbers.
Step2
Modulus of $\alpha = \left| \alpha \right| = \sqrt {{a^2} + {b^2}}$
Modulus of $\beta = \left| \beta \right| = \sqrt {{p^2} + {q^2}} = 1$
$\to {p^2} + {q^2} = 1$
Step 3
Conjugate of \alpha = \mathop {\alpha = a - ib}\limits^\\_
Step4
Now, $\left| {\dfrac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$ = $\dfrac{{\left| {\beta - \alpha } \right|}}{{\left| {1 - \overline {\alpha \beta } } \right|}}$
On putting values of $\alpha \& \beta$, we get
$= \dfrac{{\left| {(p + iq) - (a + ib)} \right|}}{{\left| {1 - (a - ib)(p + iq)} \right|}}$
$= \dfrac{{\left| {p + iq - a - ib} \right|}}{{\left| {1 - ap - aqi + bpi - bq)} \right|}}$
Separating Real and Imaginary parts,
$= \dfrac{{\left| {(p - a) + (q - b)i} \right|}}{{\left| {(1 - ap - bq)(aq - bp)i} \right|}}$
Using formula of modulus of complex number,
$= \dfrac{{\sqrt {{{(p - a)}^2} + {{(q - b)}^2}} }}{{\sqrt {{{(i - ap - bq)}^2} + {{(aq - bp)}^2}} }}$
$= \dfrac{{\sqrt {{p^2} + {a^2} - 2pa + {q^2} + {b^2} - 2qb} }}{{\sqrt {1 + {a^2}{p^2} + {b^2}{q^2} - 2ap - 2bq + 2apqb + } {a^2}{q^2} + {b^2}{q^2} - 2aqbp}}$
On combining quantities according to identities,
$= \dfrac{{\sqrt {{p^2} + {q^2} + {b^2} + {a^2} - 2pa - 2qb} }}{{\sqrt {1 + {a^2}({p^2} + {q^2}) - ({p^2} + {q^2}){b^2} - 2ap - 2bq} }}$
$\Rightarrow$ $\dfrac{{{{\left[ {1 + {a^2} + {b^2} - 2pa - 2qb} \right]}^{\dfrac{1}{2}}}}}{{{{[1 + {a^2} + {b^2} - 2ap - 2bq]}^{\dfrac{1}{2}}}}} = 1$

Note: In case of complex numbers , modulus and conjugate are the two of the main properties of complex numbers.
In step1; we have just assigned the general values of α and β .
In step 2; moduli of α and β have been calculated.
In step 3; conjugates of α are calculated.
In step 4; the values from step 1, 2& 3 are substituted.
The given values of $\left| \beta \right| = 1$ Is also substituted.
Hence, the value of $\left| {\dfrac{{\beta - \alpha }}{{1 - \overline \alpha \beta }}} \right|$ Is 1.
When α and β are complex numbers and $\left| \beta \right| = 1$.