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. 
 A: 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
$$
i^m=i^r=i^n
$$
A: Hint This image may help you understand this concept better


Edit:
First of all forget everything that was said about division by four. You may have seen $i$ being defined as $\sqrt{-1}$. We can square both sides to see an alternate definition which is $i^2=-1$. Let us use this. We first start off by noticing that anything to the power of $0$ is $1$. Therefore $$i^0=1$$ Now let's multiply each side by $i$. We get $$i\times i^0 = 1 \times i\\ i=i$$ Which is what we expect. Now we multiply by $i$ again to get $$i^2=-1$$ because we established that earlier. Now multiply by $i$ again to get $$i^3=-1\times i = -i$$ which is expected from multiplying a number by $-1$. Now finally multiply by $i$ again and watch what we get: $$i^4=-i\times i = -i^2=-(-1)=1$$ You may now be asking why this is useful well notice that if we multiply by $i$ again we will get the same cycle patter which repeats every 4! $1, i, -1, -i$. Thus $i^5$ is $i$ and $i^{10}$ is $-1$ etc. Now to work out what large power of $i$ would be we can divide the power by $4$ to see how many 4's would fit into it.
Take $i^{13}$ we can write this as $i^{12}\times i$ and we know by laws of indices/exponents that $i^{12}=(i^4)^3$. Hang on! We just established that $i^4$ is $1$ therefore $i^{12}$ is $1^3$ which is just 1! Therefore we have that $i^{13}$ is $1\times i$ which is just $i$! Notice by trying to fit as many fours into the power and rewriting them so that we can evaluate the inner part to be one and using laws of indices/exponents we can just leave a power of $i$ that is has index/exponent less than four behind.
Thus to solve the problem you need to find out what gets left behind of $2014$ when you try to fit as many fours into it. The left behind part (called the remainder) is the power of $i$ that it is equal to.
