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Question: If Z<sub>­1</sub> = \(\frac{1}{a + i}\), a ¹ 0 and Z<sub>2</sub> = \(\frac{1}{1 + bi}\) b ¹ 0 such t...

If Z­1 = 1a+i\frac{1}{a + i}, a ¹ 0 and Z2 = 11+bi\frac{1}{1 + bi} b ¹ 0 such that Z1=Zˉ2Z_{1} = {\bar{Z}}_{2} then

A

a = 1, b = 1

B

a = – 1, b = 1

C

a = 1, b = – 1

D

a = – 1, b = –1

Answer

a = 1, b = – 1

Explanation

Solution

Sol. a−ia2+1=(1−bi‾b2+1)\frac{a - i}{a^{2} + 1} = \left( \frac{\overline{1 - bi}}{b^{2} + 1} \right) Ž 1a2+1=−b1+b2\frac{1}{a^{2} + 1} = \frac{- b}{1 + b^{2}}

& aa2+1=1b2+1\frac{a}{a^{2} + 1} = \frac{1}{b^{2} + 1} Ž a = 1, b = – 1