Question
If $\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{10}=\text{a}+\text{ib},$ then find (a, b).
✓
Answer
$\text{a}+\text{ib}=\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{10}$
$=\bigg[\frac{(1-\text{i})}{(1+\text{i})}\cdot\frac{(1-\text{i})}{(1-\text{i})}\bigg]^{100}=\bigg[\frac{(1-\text{i})^2}{1-\text{i}^2}\bigg]^{100}$
$=\Big(\frac{1-2\text{i}+\text{i}^2}{1+1}\Big)^{100}=\Big(\frac{-2\text{i}}{2}\Big)^{100}$
$=(\text{i}^{4})^{25}=1$
$\therefore\ (\text{a},\text{b})=(1,0)$
$=\bigg[\frac{(1-\text{i})}{(1+\text{i})}\cdot\frac{(1-\text{i})}{(1-\text{i})}\bigg]^{100}=\bigg[\frac{(1-\text{i})^2}{1-\text{i}^2}\bigg]^{100}$
$=\Big(\frac{1-2\text{i}+\text{i}^2}{1+1}\Big)^{100}=\Big(\frac{-2\text{i}}{2}\Big)^{100}$
$=(\text{i}^{4})^{25}=1$
$\therefore\ (\text{a},\text{b})=(1,0)$
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