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If $a, b, c$ are non-zero real number 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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Verified Answer
The correct answer is:
$\frac{|\alpha|^2}{\left(1+|\alpha|^2\right)^2}$
Given that, $\alpha=\frac{a+i b}{1-c}$
$$
\Rightarrow \quad|\alpha|^2=\frac{a^2+b^2}{(1-c)^2}
$$
And $a^2+b^2+c^2=c$
$$
\Rightarrow \quad a^2+b^2=c(1-c)
$$
From Eqs. (i) and (ii),
$$
\begin{aligned}
& |\alpha|^2=\frac{c}{(1-c)} \Rightarrow \frac{1+|\alpha|^2}{1}=\frac{1}{1-c} \\
\Rightarrow \quad 1-c & =\frac{1}{1+|\alpha|^2}
\end{aligned}
$$
From Eqs. (i) and (iii), we get
$$
|\alpha|^2=\frac{a^2+b^2}{\left(\frac{1}{1+|\alpha|^2}\right)^2} \Rightarrow a^2+b^2=\frac{|\alpha|^2}{\left(1+|\alpha|^2\right)^2}
$$
$$
\Rightarrow \quad|\alpha|^2=\frac{a^2+b^2}{(1-c)^2}
$$
And $a^2+b^2+c^2=c$
$$
\Rightarrow \quad a^2+b^2=c(1-c)
$$
From Eqs. (i) and (ii),
$$
\begin{aligned}
& |\alpha|^2=\frac{c}{(1-c)} \Rightarrow \frac{1+|\alpha|^2}{1}=\frac{1}{1-c} \\
\Rightarrow \quad 1-c & =\frac{1}{1+|\alpha|^2}
\end{aligned}
$$
From Eqs. (i) and (iii), we get
$$
|\alpha|^2=\frac{a^2+b^2}{\left(\frac{1}{1+|\alpha|^2}\right)^2} \Rightarrow a^2+b^2=\frac{|\alpha|^2}{\left(1+|\alpha|^2\right)^2}
$$
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