If (1 + i)z = (1 – i) z̅,

According to the question,

We have,

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

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

Rationalizing the denominator, We get,

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

$=\frac{(1-i)^{2}}{\left(1-i^{2}\right)} \bar{z}$

$=\frac{1-2 \mathrm{i}+\mathrm{i}^{2}}{1+1} \overline{\mathrm{z}}$

$=\frac{1-2 \mathrm{i}-1}{2} \overline{\mathrm{z}}$

= -iz̅

Hence proved.