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If (1+i)z = (1-i)barz, "then show that ...

`If (1+i)z = (1-i)barz, "then show that "z = -ibarz.`

Text Solution

Verified by Experts

We have, `(1+i)z = (1-i)barz rArr barz = (barz)/(barz)=(1-i)/(1+i).`
`rArr (z)/(barz)=(1-i)/(1+i)(1-i)/(1-i) rArr (z)/(barz)=(1+i^(2)-2i)/(1-i^(2)) " " [:.i^(2)=-1] `
`rArr (z)/(barz)=(1-1-2i)/(2) rArr (z)/(barz)=i`
`rArr z = - i barz " "Hence proved. `
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Knowledge Check

  • If ((1+i) z= (1-i))bar(z), then

    A
    z lies on a straight line
    B
    for all such `z ne 0, "arg" (z) = 3pi//4`
    C
    all such z are given by `z=t (1-i), t in R`
    D
    z must be purely real
  • If (1+i)z=(1-i)barz , then z is equal to

    A
    `t(1-i), t in R`
    B
    `t(1+i),t in R`
    C
    `t/(1+i),t in R^(+)`
    D
    none of these
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