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  • If (1+i/1-i)^m=1 , then find the least positive integral value of m.

Q: 20         If   \small \left ( \frac{1+i}{1-i} \right )^m=1,   then find the least positive integral value of \small m.

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best_answer

Let 
z = \left ( \frac{1+i}{1-i} \right )^m
Now, multiply both numerator and denominator by (1+i)
We will get,
z = \left ( \frac{1+i}{1-i}\times \frac{1+i}{1+i} \right )^m
    = \left ( \frac{(1+i)^2}{1^2-i^2} \right )^m
    = \left ( \frac{1^2+i^2+2i}{1+1} \right )^m
    = \left ( \frac{1-1+2i}{2} \right )^m                                        (\because i^2 = -1)
    = \left ( \frac{2i}{2} \right )^m
    = i^m
We know that i^4 = 1
Therefore, the least positive integral value of \small m  is 4

Posted by

Gautam harsolia

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