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$i^{2014}$ 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.

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    If $n \equiv m \pmod{4}$, then $i^n = i^m$. That's because $i^4 = 1$. – Daniel Fischer Oct 25 '14 at 22:52
  • It may help to look up and understand what it means for two integers to be congruent modulo $n$ for $n>1$. ;) –  Oct 25 '14 at 23:19
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    If you've grasped the concept in the answers clearly, then you'll find my trick beneficial to solve this type of problems you face: We know that, $i^m = i^{4n + k} = i^k$ where $4n$ is the closest multiple of $4$ to $m$ and k is some integer $\in{0,1,2,3}$. Just divide $m$ by 4 and the remainder will be $k$. If you can get a hold of this idea, then we can do this effectively: $$i^{1993} = i^{1992 + 1} = i^{4\times 498 + 1} = i$$ – Nick Oct 26 '14 at 12:20
  • @Nick what about choice B? if I divide by 4 the remainder is 2. –  Oct 27 '14 at 00:27
  • @Ben: $$i^{203} = i^{200 + 3} = i^{4\times 50 + 3} = i^3 = -i$$ – Nick Oct 27 '14 at 12:16
  • @Nick so the rule is, i to any odd power will yield an imaginary number? –  Oct 27 '14 at 16:09
  • @Ben: or think of it as "remainder of an odd number when divided by 4 will always be an odd number" This is obvious isn't it? – Nick Oct 28 '14 at 02:35
  • In your question, you can simply note the fact that 2014 is even, and only 726 and 2100 are even in the options. But 2100 is double even (divisible by 4) while 2014 is only single even (2014/2 = 1007). We see that $\text{rem}_4(2014) = \text{rem}_4(726)$ – Nick Oct 28 '14 at 02:41

2 Answers2

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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 $$ i^m=i^r=i^n $$

egreg
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  • if we check which one has the remainder equal to 2014 mod 4. We can consider either option B. or C. Why isn't option B okay? 203 mod 4 gives us a remainder of 2. –  Oct 25 '14 at 23:39
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Hint This image may help you understand this concept better enter image description here

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.

Ali Caglayan
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