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# Find the least positive integer $\mathit{n}$ such that $\left ( \large\frac{1+\mathit{i}}{1-\mathit{i}} \right )^{\mathit{n}}$$= 1 Can you answer this question? ## 1 Answer 0 votes Toolbox: • If z=a+ib then , • \bar{z}=a-ib • \mid z\mid=\sqrt{a^2+b^2} • z^{-1}=\large\frac{a-ib}{a^2+b^2} • z\bar{z}=a^2+b^2 • Also Re(z)=a,Im(z)=b Step 1: \large\frac{1+i}{1-i}=\large\frac{(1+i)(1+i)}{(1-i)(1+i)} \quad\quad=\large\frac{(1+i)^2}{1+1} \quad\quad=\large\frac{1+2i+i^2}{2} \quad\quad=\large\frac{1+2i-1}{2} \quad\quad=\large\frac{2i}{2} \quad\quad=i Step 2: \big(\large\frac{1+i}{1-i}\big)^n$$=i^{\large n}$
The least positive integer n for which this value becomes 1 is $n=4$