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4-1.Complex numbers
normal
If $\left(\frac{1+i}{1-i}\right)^{m}=1$ then find the least positive integral value of $m$.
A
$4$
B
$4$
C
$4$
D
$4$
Solution
$\left(\frac{1+i}{1-i}\right)^{m}=1$
$\Rightarrow\left(\frac{1+i}{1-i} \times \frac{1+i}{1+i}\right)^{m}=1$
$\Rightarrow\left(\frac{(1+i)^{2}}{1^{2}+1^{2}}\right)^{m}=1$
$\Rightarrow\left(\frac{1^{2}+i^{2}+2 i}{2}\right)^{m}=1$
$\Rightarrow\left(\frac{1-1+2 i}{2}\right)^{m}=1$
$\Rightarrow\left(\frac{2 i}{2}\right)^{m}=1$
$\Rightarrow i^{m}=1$
$\Rightarrow i^{m}=i^{4 k}$
$\therefore m=4 k,$ where $k$ is some integer.
Therefore, the least positive integer is $1 .$
Thus, the least positive integral value of $m$ is $4(=4 \times 1)$
Standard 11
Mathematics
