Question

If α and β are different complex numbers with | β |= 1, then find | (β-α)/(1- αβ) |.

Answer 1

Given:     α and β are different complex numbers with | β |= 1,

To find :   | (β-α)/(1- αβ) |.  correction  $| \frac{\beta - \alpha }{1 - \overline\alpha \beta} |$

Solution:

α and β are different complex numbers

α = a + ib

β = c + id

| β |= 1,  => c² + d²  = 1

|(β-α)/(1- αβ) |.

= |   ((c + id) - (a + ib)) / ( 1  - (a + ib)(c + id) ) |

= | ( (c - a) + i(d - b ) ) / ( 1  - (ac - bd + ibc + iad ) ) |

=  | ( (c - a) + i(d - b ) ) / ( (1  -  ac +  bd) + i(bc + ad ) ) |

as | z1 / z2| = |z1| /| z2|

=  √ (c - a)² + (d - b)² / √( (1 - ac + bd)² + (bc + ad)²)

=  √( c² + a² - 2ac + d² + b² - 2bd) /  / √(1 + a²c² + b²d² -2ac + 2bd - 2abcd + b²c² + a²d²  - 2abcd)

using c² + d² = 1

= √(  1 + a² + b²- 2ac - 2bd) /  / √(1 + a²(c² + d² ) + b²(c² + d² ) -2ac + 2bd  )

= √(  1 + a² + b²- 2ac - 2bd) /  / √(1 + a²  + b² -2ac + 2bd  )

Difficult to find exact value

Here is correction in Question

Correction   $| \frac{\beta - \alpha }{1 - \overline\alpha \beta} |$

= |   ((c + id) - (a + ib)) / ( 1  - (a - ib)(c + id) ) |

= | ( (c - a) + i(d - b ) ) / ( 1  - (ac + bd - ibc + iad ) ) |

=  | ( (c - a) + i(d - b ) ) / ( (1  -  ac  -  bd) + i(bc - ad ) ) |

as | z1 / z2| = |z1| /| z2|

=  √ (c - a)² + (d - b)² / √( (1 - ac - bd)² + (bc - ad)²)

= √( c² + a² - 2ac + d² + b² - 2bd) /  / √(1 + a²c² + b²d² -2ac - 2bd + 2abcd + b²c² + a²d²  - 2abcd)

using c² + d² = 1

= √(  1 + a² + b²- 2ac - 2bd) /  / √(1 + a²(c² + d² ) + b²(c² + d² ) -2ac - 2bd  )

= √(  1 + a² + b²- 2ac - 2bd) /  / √(1 + a² + b² -2ac - 2bd  )

= 1

 $| \frac{\beta - \alpha }{1 - \overline\alpha \beta} |$ = 1

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