Question

Express the mentioned below complex number in the standard form $a+ib$
(iv) $\frac{1-i}{1+i}$

Answer 1

$\because \frac{1-i}{1+i}$

$=\frac{1-i}{1+i}\times \frac{1-i}{1-i}$

$=\frac{(1-i{)}^2}{1^2-i^2}$

$=\frac{1+i^2-2i}{1-(-1)}$

$=\frac{-2i}{2}$

$=-i$

$\therefore \frac{1-i}{1+i}=-i$