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Question
If \[\left( \frac{1 - i}{1 + i} \right)^{100} = a + ib\] find (a, b).
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Solution
\[\frac{1 - i}{1 + i} = \frac{1 - i}{1 + i} \times \frac{1 - i}{1 - i}\]
\[ = \frac{\left( 1 - i \right)^2}{1^2 - i^2}\]
\[ = \frac{1^2 + i^2 - 2i}{1 + 1} [ \because i^2 = - 1] \]
\[ = \frac{1 - 1 - 2i}{2}\]
\[ = \frac{- 2i}{2}\]
\[ = - i . . . . (1)\]
It is given that,
\[\left( \frac{1 - i}{1 + i} \right)^{100} = a + ib\]
\[ \Rightarrow ( - i )^{100} = a + ib [\text { From } (1)]\]
\[ \Rightarrow i^{4 \times 25} = a + ib\]
\[ \Rightarrow 1 + 0i = a + ib [ \because i^4 = 1]\]
\[ \Rightarrow a = 1 \text { and } b = 0\]
Thus, (a, b) = (1, 0).
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