Question:
If $(1+i) z=(1-i)^{\bar{z}}$ then prove that $z=-i \bar{z}$.
Solution:
Let z = x + iy
Then
$\bar{z}=x-i y$
Now, Given: $(1+i) z=(1-i) \bar{z}$
Therefore,
$(1+i)(x+i y)=(1-i)(x-i y)$
$x+i y+x i+i^{2} y=x-i y-x i+i^{2} y$
We know that $i^{2}=-1$, therefore,
$x+i y+i x-y=x-i y-i x-y$
$2 x i+2 y i=0$
$x=-y$
Now, as $x=-y$
$z=-\bar{Z}$
Hence, Proved.