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If \( \left(\frac{1+i}{1-i}\right)^{\mathrm{m}} \), then the least positive integralvalue of \( \mathrm{m} \) is
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The correct answer is:
\( 04 \)
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, $m=4$
$\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$
$\Rightarrow\left(\frac{(1+i)^{2}}{1-(i)^{2}}\right)^{m}=1$
$\Rightarrow\left(\frac{1-1+2 i}{1+1}\right)^{m}=1$
$\Rightarrow\left(\frac{2 i}{2}\right)^{m}=1$
$\Rightarrow i^{m}=1$
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, $m=4$
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