Rules of i ($\sqrt -1$) to a power $i^{2014}$ power =?
A. $i^{13}$
B. $ i ^{203}$
C. $i^{726}$
D. $i^{1993}$
E. $i^{2100}$
I don't understand the concept that powers of i repeat in fours and that "two powers of i are equal if their remainders are equal upon division by four". I especially don't understand the second part of the previous statement. 

You know that $i^4=1$, because
$$
i^4=(i^2)^2=(-1)^2=1
$$
Now $i^5=i^4\cdot i=i$, $i^6=i^4\cdot i^2=-1$, $i^7=i^4\cdot i^3=-i$ and finally $i^8=i^4\cdot i^4=1$. You can go on forever, the powers of $i$ will repeat the same pattern
$$\dots\quad i\quad {-1} \quad {-i}\quad 1 \quad\dots$$
If $n=4q+r$ with $0\le r<4$, that is, $r$ is the remainder of the division of $n$ by $4$, you have
$$
i^n=i^{4q+r}=i^{4q}\cdot i^r=(i^4)^q\cdot i^r=1^q\cdot i^r=i^r
$$
In particular, if $m=4q_1+r$ and $n=4q_2+r$ have the same remainder, then
$$
i^m=i^r=i^n
$$
