If z1=a+i1,a=0 and z2=1+bi1, b=0 are such that z1=zˉ2 then

Given z1=a+i1=a+i1a−ia−i=1+a2a−iand z2=1+bi1=1+bi11−bi1−bi=1+b21−biAlso given, z1=z2ˉ⇒1+a2a−i=1+b21+bi⇒1+a2a=1+b21 and 1+a2−1=1+b2bBy solving these equation we get a=1,b=−1

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If z_1 = {1}{a + i}, a 0 and z_2 ={1}{1 + bi} , b 0 are

Question

If $z_{1} = \frac{1}{a + i}, a \neq = 0$ and $z_{2} = \frac{1}{1 + bi}, b \neq = 0$ are such that $z_{1} = \overset{z}{ˉ}_{2}$ then

a = - 1, b = 1

a = 1, b = - 1

a = 2, b = 1

a = 1, b = 2

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Mateo Discussed

z_{1} = \frac{1}{a + i}, a \neq = 0

and

z_{2} = \frac{1}{1 + bi}, b \neq = 0

are such that

z_{1} = \overset{z}{ˉ}_{2}

then

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Text solutionVerified

Given

z_{1} = \frac{1}{a + i} = \frac{1}{a + i} \frac{a - i}{a - i} = \frac{a - i}{1 + a ^{2}}

and

z_{2} = \frac{1}{1 + bi} = \frac{1}{1 + bi} \frac{1 - bi}{1 - bi} = \frac{1 - bi}{1 + b ^{2}}

Also given,

z_{1} = \overset{z_{2}}{ˉ}

\Rightarrow \frac{a - i}{1 + a ^{2}} = \frac{1 + bi}{1 + b ^{2}}

\Rightarrow \frac{a}{1 + a ^{2}} = \frac{1}{1 + b ^{2}}

and

\frac{- 1}{1 + a ^{2}} = \frac{b}{1 + b ^{2}}

By solving these equation we get

a = 1, b = - 1

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Question Text	If $z_{1} = \frac{1}{a + i}, a \neq = 0$ and $z_{2} = \frac{1}{1 + bi}, b \neq = 0$ are such that $z_{1} = \overset{z}{ˉ}_{2}$ then
Answer Type	Text solution:1
Upvotes	150

If z1​=a+i1​,a=0 and z2​=1+bi1​, b=0 are such that z1​=zˉ2​ then

Text solutionVerified

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If $z_{1} = \frac{1}{a + i}, a \neq = 0$ and $z_{2} = \frac{1}{1 + bi}, b \neq = 0$ are such that $z_{1} = \overset{z}{ˉ}_{2}$ then