if "i" is raised to an odd power, then it cannot simplify to be
a)-1
b)-i
c)i

Question

if "i" is raised to an odd power, then it cannot simplify to be 
a)-1
b)-i
c)i

anonymous · Answer

-1

lgbasallote · Answer

$$i^1 = i$$
$$i^2 = -1$$
$$i^3 = i^2 \times i = -1 \times i = -i$$
$$i^4 = (i^2)^2 = (-1)^2 = 1$$
\[i^5 = (i^2)^2 \times i = 1 \times i = i]
the process repeats

lgbasallote · Answer

$$i^5 = (i^2)^2 \times i = 1 \times i = i$$

jamesj · Answer

@sauravshakya, please do not give answers with no explanation. It is contrary to the ethos of this site and against our code of conduct: http://openstudy.com/code-of-conduct

klimenkov · Answer

Odd number $n$ is a number that can be written in form 
$n=2k-1, \ k$ is integer.
$i^{2k-1}=i^{2k}\cdot i^{-1}$
$i^{2k}=(i^2)^k=(-1)^k$
$i^{-1}=\frac 1 i=\frac {i} {i\cdot i}=-i$
So, $i^{2k-1}=i^{2k}\cdot i^{-1}=(-1)^k\cdot (-i)=(-1)^{k+1}\cdot i$
For any integer $k$ it will be a complex number, but not an integer.

anonymous · Answer

ok. My mistake