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Question
If `((1 - i)/(1 + i))^100` = a + ib, then find (a, b).
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Solution
We have `((1 - i)/(1 + i))^100` = a + bi
⇒ `((1 - i)/(1 + i) xx (1 - i)/(1 - i))^100` = a + bi
⇒ `((1 + i^2 - 2i)/(1 - i^2))^100` = a + bi
⇒ `((1 - 1 - 2i)/(1 + 1))^100` = a + bi
⇒ `((-2i)/2)^100` = a + bi
⇒ (–i)100 = a + bi
⇒ i100 = a + bi
⇒ (i4)25 = a + bi
⇒ (1)25 = a + bi
⇒ 1 = a + bi
⇒ 1 + 0i = a + bi
Comparing the real and imaginary parts,
We have a = 1, b = 0
Hence (a, b) = (1, 0)
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