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4-1.Complex numbers
medium
જો $n$ એ ધન પૂર્ણાક હોય, તો ${\left( {\frac{{1 + i}}{{1 - i}}} \right)^{4n + 1}}$=
A
$1$
B
$-1$
C
$i$
D
$ - i$
Solution
(c)Since $\frac{{1 + i}}{{1 – i}} = \frac{{(1 + i)(1 + i)}}{{(1 – i)(1 + i)}} = i$
Therefore ${\left( {\frac{{1 + i}}{{1 – i}}} \right)^{4n + 1}} = {i^{4n + 1}} = i{i^{4n}} = i\,\,\,\,\,({i^{4n}} = 1)$.
Std 11
Mathematics
