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If $\;\alpha\;$ and $\;\beta\;$ are different complex numbers with $\;|\beta=1|\;$ , then find $\;|\large\frac{\beta-\alpha}{1-\overline{\alpha} \beta}|$

$(a)\;1\qquad(b)\;0\qquad(c)\;2\qquad(d)\;4$

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Answer : 1
Explanation :
Let $\; \alpha = a+ib\;$ and $\; \beta = x+iy$
It is given that $\;|\beta| =1$
Therefore , $\; \sqrt{x^{2}+y^{2}} =1$
$x^{2} + y^{2} =1$ ----(i)
$\;|\large\frac{\beta-\alpha}{1-\overline{\alpha} \beta}| = | \large\frac{(x+iy)-(a+ib)}{(1-(a-ib)(x+iy)}|$
$ = |\large\frac{(x-a)+i(y-b)}{1-(ax+aiy-ibx-i^{2}by)}|$
$ = |\large\frac{(x-a)+i(y-b)}{(1-ax-by)+ i (bx-ay)}| \qquad |\large\frac{z_{1}}{z_{2}}| = \large\frac{|z_{1}|}{|z_{2}|}$
$= \large\frac{\sqrt{(x-a)^{2}+(y-b)^{2}}}{\sqrt{(1-ax-by)^{2}+(bx-ay)^{2}}}$
$= \large\frac{\sqrt{x^{2}+a^{2}-2ax+y^{2}+b^{2}-2by}}{\sqrt{1^{2}+a^{2}x^{2}+b^{2}y^{2}-2ax-2abxy-2by+b^{2}x^{2}+a^{2}y^{2}-2abxy}}$
$= \large\frac{\sqrt{(x^{2}+y^{2})+a^{2}+b^{2}-2ax-2by}}{\sqrt{1+a^{2}(x^{2}+y^{2})+b^{2}(y^{2}+x^{2})-2ax-2by}}$
$= \large\frac{\sqrt{1+a^{2}+b^{2}-2ax-2by}}{\sqrt{1+a^{2}+b^{2}-2ax-2by}}$
$=1$
$\;|\large\frac{\beta-\alpha}{1-\overline{\alpha} \beta}| =1$
answered Apr 15, 2014 by yamini.v
edited Apr 15, 2014 by yamini.v
 

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