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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$

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1 Answer

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 $n=4$
answered Jun 7, 2013 by sreemathi.v
 

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