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Question
If (a + ib) = `(1 + "i")/(1 - "i")`, then prove that (a2 + b2) = 1
Solution
(a + ib) = `(1 + "i")/(1 - "i") = (1 + "i")^2/((1 - "i")(1 + "i"))`
∴ a + bi = `(1 + 2"i" + "i"^2)/(1 - "i"^2)`
= `(1 + 2"i" - 1)/(1 - (- 1))` ...[∵ i2 = – 1]
= `(2"i")/2`
= i
∴ a + bi = 0 + i
Equating real and imaginary parts, we get
a = 0 and b = 1
∴ a2 + b2 = 02 + 12 = 1
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