If

$(1+i) z=(1-i) \bar{z}$

$\Rightarrow \frac{z}{\bar{z}}=\frac{1-i}{1+i}$

$\Rightarrow \frac{z}{\bar{z}}=\frac{1-i}{1+i} \times \frac{1-i}{1-i}$

$\Rightarrow \frac{z}{\bar{z}}=\frac{1+i^{2}-2 i}{1-i^{2}}$

$\Rightarrow \frac{z}{\bar{z}}=\frac{1-1-2 i}{1+1}$   $\left[\because i^{2}=-1\right]$

$\Rightarrow \frac{z}{\bar{z}}=\frac{-2 i}{2}$

$\Rightarrow \frac{z}{\bar{z}}=-i$

$\Rightarrow z=-i \bar{z}$

Hence, $z=-i \bar{z}$.