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Question
If (1 + i)z = `(1 - i)barz`, then show that z = `-ibarz`.
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Solution
Given that: (1 + i)z = `(1 - i)barz`
⇒ `z/barz = (1 - i)/(1 + i)` = `(1 - i)/(1 + i) xx (1 - i)/(1 - i)`
= `(1 + i^2 - 2i)/(1 - i^2)` = `(1 - 1 - 2i)/(1 + 1)`
= `(-2i)/2` = –i
⇒ `z/barz` = –i
∴ z = `-i barz`
Hence proved.
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