Question

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

• A. $-1$
• B. $1$
• C. $-i$
• D. $i$

Answer 1

Answer: (C)

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