Question:
The conjugate of the complex number $\frac{1-i}{1+i}$ is
Solution:
$\frac{(\overline{1-i})}{(\overline{1+i})}$ i. e conjugate of $\frac{1-i}{1+i}$
$=\frac{(\overline{1-i})}{(\overline{1+i})}$
$=\frac{1+i}{1-i}$
$=\frac{1+i}{1-i} \times \frac{1+i}{1+i}$
$=\frac{(1+i)^{2}}{1-i^{2}}$
$=\frac{(1+i)^{2}}{1+1}=\frac{1}{2}\left(1+i^{2}+2 i\right)$
Conjugate of $\frac{1-i}{1+i}=\frac{1}{2}(2 i)=i$
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