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If $\;(\large\frac{1+i}{1-i})^{m}=1\;$ then find the least positive integral value of m

$(a)\;4\qquad(b)\;2\qquad(c)\;0\qquad(d)\;1$

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Answer : 4
Explanation :
$\;(\large\frac{1+i}{1-i})^{m}=1\; $
$ (\large\frac{1+i}{1-i} \times \large\frac{1+i}{1+i})^{m}=1$
$ (\large\frac{(1+i)^{2}}{1^{2} - i^{2}})^{m}=1$
$(\large\frac{1^{2}+i^{2}+2i}{2})^{m}=1$
$(\large\frac{2i}{2})^{m}=1$
$i^{m}=1$
Therefore m =4k where k is some integer
Therefore , the least positive integer is 1 .
Thus , the least positive integral value of m is 4 =$\; (4 \times 1)$
answered Apr 15, 2014 by yamini.v
 

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