If a, b, c are non-zero real numbers with c n | vedclass

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(A) Given that,$\alpha = \frac{a+ib}{1-c}$.Taking the modulus squared on both sides,we get $|\alpha|^2 = \frac{a^2+b^2}{(1-c)^2} \quad \dots(i)$.Given

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$\frac{|\alpha|^2}{(1+|\alpha|^2)^2}$

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(A) Given that,$\alpha = \frac{a+ib}{1-c}$.Taking the modulus squared on both sides,we get $|\alpha|^2 = \frac{a^2+b^2}{(1-c)^2} \quad \dots(i)$.Given $a^2+b^2+c^2=c$,we have $a^2+b^2 = c-c^2 = c(1-c) \quad \dots(ii)$.Substituting $(ii)$ into $(i)$,we get $|\alpha|^2 = \frac{c(1-c)}{(1-c)^2} = \frac{c}{1-c}$.This implies $1+|\alpha|^2 = 1 + \frac{c}{1-c} = \frac{1-c+c}{1-c} = \frac{1}{1-c}$.Thus,$1-c = \frac{1}{1+|\alpha|^2}$.From $(ii)$,$a^2+b^2 = c(1-c) = (1-(1-c))(1-c) = (1-c) - (1-c)^2$.Substituting $1-c = \frac{1}{1+|\alpha|^2}$,we get $a^2+b^2 = \frac{1}{1+|\alpha|^2} - \left(\frac{1}{1+|\alpha|^2}\right)^2 = \frac{1+|\alpha|^2-1}{(1+|\alpha|^2)^2} = \frac{|\alpha|^2}{(1+|\alpha|^2)^2}$.

If $a, b, c$ are non-zero real numbers with $c \neq 1$ such that $a^2+b^2+c^2=c$ and if $\alpha=\frac{a+i b}{1-c}$,then $a^2+b^2=$

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