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4-1.Complex numbers
medium
If ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^x} = 1,$ then
A
$x = 4n$, where $n$ is any positive integer
B
$x = 2n$, where $n$ is any positive integer
C
$x = 4n + 1$, where $n$ is any positive integer
D
$x = 2n + 1$, where $n$ is any positive integer
(AIEEE-2003)
Solution
(a) ${\left( {\frac{{1 + i}}{{1 – i}}} \right)^x} = 1,$
==> ${\left( {\frac{{1 + {i^2} + 2i}}{{1 + 1}}} \right)^x} = 1\,\, \Rightarrow \,{i^x} = 1\,$
$\therefore x = 4n,\,n \in {I^ + }$.
Standard 11
Mathematics
