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AP EAMCET · Maths · Complex Number

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 \(a^2+b^2=\)

  1. A \(\frac{|\alpha|^2}{\left(1+|\alpha|^2\right)^2}\)
  2. B \(\frac{|\alpha|^4}{\left(1+|\alpha|^2\right)^2}\)
  3. C \(\frac{|\alpha|}{1+|\alpha|^2}\)
  4. D \(\frac{|\alpha|}{1+|\alpha|}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{|\alpha|^2}{\left(1+|\alpha|^2\right)^2}\)

Step-by-step Solution

Detailed explanation

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),…