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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

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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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