Q64 / 180 · KCET 2017 · Mathematics

If \( \left(\frac{1+i}{1-i}\right)^{\mathrm{m}} \), then the least positive integralvalue of \( \mathrm{m} \) is

  • A
    \( 02 \)
  • B
    \( 13 \)
  • Correct answer
    C
    \( 04 \)
  • D
    \( 11 \)

Solution

Given that, $\left(\frac{1+i}{1-i}\right)^{m}=1$ $\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$ $\Rightarrow\left(\frac{(1+i)^{2}}{1-(i)^{2}}\right)^{m}=1$ $\Rightarrow\left(\frac{1-1+2 i}{1+1}\right)^{m}=1$ $\Rightarrow\left(\frac{2 i}{2}\right)^{m}=1$ $\Rightarrow i^{m}=1$ For $m=1$, we have $i=1$, but $i \neq 1$ For $m=2$, we have $i^{2}=-1=1$, but $-1 \neq 1$ For $m=3$, we have $i^{3}=-i=1$, but $-i \neq 1$ For $m=4$, we have $i^{4}=1$ Therefore, $m=4$

Hence, the correct answer is option: 'C'.

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