Question
Find the modulus of $\frac{1+\text{i}}{1-\text{i}}-\frac{1-\text{i}}{1+\text{i}}$

Answer

Let $\text{z}=\frac{1+\text{i}}{1-\text{i}}-\frac{1-\text{i}}{1+\text{i}}$ $=\frac{(1+\text{i})^2-(1-\text{i})^2}{(1-\text{i})(1+\text{i})}$ $=\frac{1^2+\text{i}^2+2\times1\times\text{i}-(1^2+\text{i}^2-2\times1\times\text{i})}{1^2+1^2}$ $=\frac{1-1+2\text{i}-(1-1-2\text{i})}{2}$ $=\frac{2\text{i}+2\text{i}}{2}$ $=\frac{4\text{i}}{2}$ $\Rightarrow\text{z}=2\text{i}$ $\therefore \ |\text{z}|=|2\text{i}|$ $=2|\text{i}| \ \Big(\because \ |\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|\Big)$ $=2\times\text{i} \ \Big(\because \ |\text{i}|=1\Big)$ $=2$

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