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  • If α and β are different complex numbers with |β| =1, then find |β-α/1- conjugate α β|

Q : 17        If \small \alpha and \small \beta are different complex numbers with  \small |\beta|=1 , then find          

                  \small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |. 
 

Answers (1)

best_answer

Let
\alpha = a+ib    and      \beta = x+iy
It is given that
\small |\beta|=1\Rightarrow \sqrt{x^2+y^2} = 1\Rightarrow x^2+y^2 = 1
and 
\small \bar \alpha = a-ib
Now,
\small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right | = \left | \frac{(x+iy)-(a+ib)}{1-(a-ib)(x+iy)} \right | = \left | \frac{(x-a)+i(y-b)}{1-(ax+iay-ibx-i^2yb)} \right |
                                                                    \small = \left | \frac{(x-a)+i(y-b)}{(1-ax-yb)-i(bx-ay)} \right |
                                                                   \small = \frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-ax-yb)^2+(bx-ay)^2}}
                                                                  \small = \frac{\sqrt{x^2+a^2-2xa+y^2+b^2-yb}}{\sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-by+b^2x^2+a^2y^2-2abxy}}
                                                                   \small = \frac{\sqrt{(x^2+y^2)+a^2-2xa+b^2-yb}}{\sqrt{1+a^2(x^2+y^2)+b^2(x^2+y^2)-2ax+2abxy-by-2abxy}}
                                                                  \small = \frac{\sqrt{1+a^2-2xa+b^2-yb}}{\sqrt{1+a^2+b^2-2ax-by}}                                    \small (\because x^2+y^2 = 1 \ given)
                                                                  \small =1
                                                                   
Therefore, value of   \small \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |   is  1
 

Posted by

Gautam harsolia

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