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If (a + ib) = 1+i1-i, then prove that (a2 + b2) = 1

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Question

If (a + ib) = `(1 + "i")/(1 - "i")`, then prove that (a2 + b2) = 1

Sum
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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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Chapter 1: Complex Numbers - Exercise 1.1 [Page 6]

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