1.

if `(1+i)z=(1-i)barz` then `z` is

Answer» `(1+i)z=(1-i)bar(z) rArr (z)/(bar(z))=((1-i))/((1+i))xx((1-i))/((1-i))=((1-i)^(2))/(2)=((1+i^(2)-2i))/(2)= -i`.
Hence, `z = -i bar(z)`.


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