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…
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 $a^2+b^2=$
- $\frac{|\alpha|^2}{\left(1+|\alpha|^2\right)^2}$
- $\frac{|\alpha|^4}{\left(1+|\alpha|^2\right)^2}$
- $\frac{|\alpha|}{1+|\alpha|^2}$
- $\frac{|\alpha|}{1+|\alpha|}$
Solution
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}
$
Asked in: AP EAMCET 2018 (23 Apr Shift 2)
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