If {left( {frac{{1 + i}}{{1 - i}}} right)^m} = 1, then least integral value of m is . . . .

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4-1.Complex numbers

medium

If ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^m} = 1,$then the least integral value of $m$ is . . . .

$2$

$4$

$8$

None of these

(IIT-1982)

Solution

(b)$\frac{{1 + i}}{{1 – i}} = \frac{{1 + i}}{{1 – i}} \times \frac{{1 + i}}{{1 + i}} = \frac{{{{(1 + i)}^2}}}{2} = \frac{{2i}}{2} = i$
$\therefore $${\left( {\frac{{1 + i}}{{1 – i}}} \right)^m} = {i^m} = 1$(as given)
So the least value of $m = 4$.

$\{ \because {i^4} = 1\} $

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If ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^m} = 1,$then the least integral value of $m$ is . . . .

Solution

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