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Q. If $k = 4n + 7$, then $i^k$ equals $(n \in I)$

Complex Numbers and Quadratic Equations

Solution:

$i^k = i^{4n} i^7 = (i^{4})^n (i^2)^3i = 1 (- 1) (i) = - i$
Alternative Solution:
$i^k = i^{4n +7} = i^{4n+ 4} + i^3$
$ = (i^4)^{n + 1} i^2i = 1(-1)i = -i$