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Question
If \[\frac{1 - ix}{1 + ix} = a + ib\] then \[a^2 + b^2\]
Options
1
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0
none of these
Solution
1
\[\frac{1 - ix}{1 + ix} = a + ib\]
\[\text { Taking modulus on both the sides, we get }: \]
\[\left| \frac{1 - ix}{1 + ix} \right| = \left| a + ib \right|\]
\[ \Rightarrow \frac{\sqrt{1^2 + x^2}}{\sqrt{1^2 + x^2}} = \sqrt{a^2 + b^2}\]
\[ \Rightarrow \sqrt{a^2 + b^2} = 1\]
\[\text { Squaring both the sides, we get: } \]
\[ a^2 + b^2 = 1\]
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