# The conjugate of the complex number $\;\large\frac{1-i}{1+i}\;$ is

$(a)\;-2i\qquad(b)\;-i\qquad(c)\;i\qquad(d)\;4i$

Explanation :
$\large\frac{1-i}{1+i} = \large\frac{1-i}{1+i} \times \large\frac{1-i}{1-i}$
$= \large\frac{(1-i)^{2}}{1^{2}-i^{2}}$
$= \large\frac{1^{2}+i^{2} - 2i}{1+1}$
$= \large\frac{1-1-2i}{2}$
$= -i$
Hence , conjugate of $\;\large\frac{1-i}{1+i}\;$ is i .