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KCET 2017 Math Solved Paper

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Question : 4 of 60

Marks: +1, -0

If

\left(\frac{1+i}{1-i}\right)^m

,then the least positive integralvalue of m is

2
3
4
1

Solution:

Given that,

\left(\; \frac{1+i}{1-i}\right)^{m}=1

\Rightarrow \left( \; \frac{1+i}{1-i} \times \; \frac{1+i}{1+i} \right)^{m}=1

\Rightarrow \left( \; \frac{(1+i)^{2}}{1-i^{2}} \right)^{m}=1

\Rightarrow \left( \; \frac{1-1+2i}{1+1} \right)^{m}=1

\Rightarrow \left( \; \frac{2i}{2} \right)^{m}=1

\Rightarrow i^{m}=1

For m = 1, we have i = 1, but i ≠ 1 .

For

m = 2

, we have

i^2 = -1

= 1, but −1 ≠ 1 .

For m = 3, we have i

i^3 = -i = 1

, but −i ≠ 1 .

For m = 4, we have

i^4 = 1

.

Therefore, m = 4

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