Question:
Find the least positive integer n for which $\left(\frac{1+i}{1-i}\right)^{n}=1$
Solution:
We have, $\left(\frac{1+i}{1-i}\right)^{n}=1$
Now, $\frac{1+\mathrm{i}}{1-\mathrm{i}}=\frac{1+\mathrm{i}}{1-\mathrm{i}} \times \frac{1+\mathrm{i}}{1+\mathrm{i}}$
$=\frac{(1+i)^{2}}{1^{2}-i^{2}}$
$=\frac{1^{2}+2 i+i^{2}}{1-(-1)}$
$=\frac{1+2 i-1}{2}$
$=i$
$\therefore\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}=1 \Rightarrow \mathrm{n}$ is multiple of 4
$\therefore$ The least positive integer $\mathrm{n}$ is 4
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