Question
Express the following complex numbers in the standard form a + ib: $\frac{1-\text{i}}{1+\text{i}}$

Answer

$\frac{1-\text{i}}{1+\text{i}}=\frac{(1-\text{i})}{(1+\text{i})}\times\frac{(1-\text{i})}{(1-\text{i})}$ [On rationalising the denominator] $=\frac{(1-\text{i})^2}{1^2+1^2} \ \big[\therefore \ (\text{a}+\text{ib})(\text{a}-\text{ib})=\text{a}^2+\text{b}^2\big]$ $=\frac{1^2+\text{i}^2-2\times\text{i}\times1}{2}$ $=\frac{-2\text{i}}{2}$ $=\text{-i}$ $=0-\text{i}$ $\therefore \ \frac{1-\text{i}}{1+\text{i}}=0-\text{i}$

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