# Does $i^4$ equal $1?$

I can't seem to find a solution to this for the life of me. My mathematics teacher didn't know either.

Edit: I asked the teacher that usually teaches my course today, and she said it was incredible that the other teacher didn't know.

My logic goes as follows:

any real number: $x$ to the fourth power is equal to $(x^2)^2$. Using this logic, $i^4$ would be equal to $(i^2)^2$. This would result in $(-1)^2$, and $(-1)^2 = 1$.

Obviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?

• Yes, $i^4=1$. But what else could it equal ?? – Yves Daoust Jan 7 '17 at 15:00
• To check if $i^4$ is real or not, why not try conjugating it and see what happens? – J. M. isn't a mathematician Jan 7 '17 at 15:11
• Yes, @J.M.isn'tamathematician, but before you proceed to prove something, I'm going to get some paper and work it out for myself – Travis Jan 7 '17 at 15:21
• "I'm going to get some paper and work it out for myself" - that was my intention behind asking you those questions. – J. M. isn't a mathematician Jan 7 '17 at 15:24
• If you're working within the naive approach to complex numbers often used when they're first introduced the students, "Suppose $i^2=-1$ but that other than that all the usual rules of arithmetic apply", then basically the bottom line is that this is one of the "usual rules" which is simply assumed to apply. Specifically, we assume that complex numbers are associative, that is that for any $a, b, c$, $a(bc)=(ab)c$. You might try showing in detail why this assumption implies $i^4=1$. – Jack M Jan 12 '17 at 21:03

Yes. The powers of $i$ are cyclic, repeating themselves ever time the exponent increases by 4: $$i^0 = 1$$ $$i^1=i$$ $$i^2 = -1$$ $$i^3 = -i$$ $$i^4 = 1$$ $$i^5 = i$$ $$i^6 = -1$$ $$i^7 = -i$$ $$i^8 = 1$$ etc.

Your reasoning is excellent, and you should feel good about the fact that you figured this out on your own. The fact that your math teacher didn't know this is, in my professional opinion as a mathematics educator, a disgrace.

Edited to add: As Kamil Maciorowski notes in the comments, the pattern persists for negative exponents, as well. Specifically, $$i^{-1}= \frac{1}{i} = -i$$ If $\frac{1}{i}=-i$ seems odd, notice that $i(-i) = -i^2 = -(-1) = 1$, so $i$ and $-i$ are multiplicative inverses; therefore $i^{-1} = -i$. Once you know that, you can extend the pattern: $$i^{-1} = -i$$ $$i^{-2} = -1$$ $$i^{-3} = i$$ $$i^{-4} = 1$$ and so on.

Second update: The OP asks for some additional discussion of the property $\left( x^a \right)^b = x^{ab}$, so here is some background on that:

First, if $a$ and $b$ are natural numbers, then exponentiation is most naturally understood in terms of repeated multiplication. In this context, $x^a$ means $(x\cdot x\cdot \cdots \cdot x)$ (with $a$ factors of $x$ appearing), and $\left( x^a \right)^b$ means $(x\cdot x\cdot \cdots \cdot x)\cdot(x\cdot x\cdot \cdots \cdot x)\cdot \cdots \cdot (x\cdot x\cdot \cdots \cdot x)$, with $b$ sets of parentheses, each containing $a$ factors of $x$. Since multiplication is associative, we can drop the parentheses and recognize this as a product of $ab$ factors of $x$, i.e. $x^{ab}$.

Note that this reasoning works for any $x$, whether it is positive, negative, or complex. It even applies in settings were multiplication is noncommutative, like matrix multiplication or quaternions. All we need is that multiplication is associative, and that $a$ and $b$ be natural numbers.

Once we have established that $\left( x^a \right)^b = x^{ab}$ for natural numbers $a,b$ we can extend the logic to integer exponents. If $a$ is a positive number, and if $x$ has a multiplicative inverse, then we define $x^{-a}$ to mean the same thing as $\left(\frac1x\right)^a$, or (equivalently) as $\frac1{x^a}$. With this convention in place, it is straightforward to verify that for any combination of signs for $a,b$, the formula $\left(x^a\right)^b = x^{ab}$ holds.

Note however that in extending the formula to cover a larger set of exponents, we have also made it necessary to restrict the domain of values $x$ over which this property holds. If $a$ and $b$ are just natural numbers then $x$ can be almost any object in any set over which an associative multiplication is defined. But if we want to allow $a$ and $b$ to be integers then we have to restrict the formula to the case where $x$ is an invertible element. In particular, the formula $x^{a}$ is not really well-defined if $x=0$ and $a$ is negative.

Now let's consider the case where the exponents are not just integers but arbitrary rational numbers. We begin by defining $x^{1/a}$ to mean $\sqrt[a]{x}$. ( See Why does $x^{\frac{1}{a}} = \sqrt[a]{x}$? for a short explanation of why this convention makes sense.)

In this definition, we are assuming that $a$ is a natural number, and that $x$ is positive. Why do we need $x$ to be positive? Well, consider an expression like $x^{1/2}$. If $x$ is positive, this is (by convention) defined to be the positive square root of $x$. But if $x$ is negative, then $x^{1/2}$ is not a real number, and even if we extend our number system to include complex numbers, it is not completely clear which of the two complex square roots of $x$ this should be identified with. More or less the same problem arises when you try to extend the property to complex $x$: while nonzero complex numbers do have square roots (and $n$th roots in general), there is no way to choose a "principal" $n$th root.

Things get really crazy when you try to extend the property $\left(x^a\right)^b=x^{ab}$ to irrational exponents. If $x$ is a positive real number and $a$ is a real number, we can re-define the expression $x^a$ to mean $e^{a\ln x}$, and it can be proved that this re-definition produces the same results as all of the conventions above, but it only works because $\ln x$ is well-defined for positive $x$. As soon as you try to allow negative $x$, you run into trouble, since $\ln x$ isn't well-defined in that case. One can define logarithms of negative and complex numbers, but they are not single-valued, and there are all kinds of technicalities about choosing a "branch" of the logarithm function.

In particular -- and this is very important for the question at hand -- the identity $\left(x^a\right)^b=x^{ab}$ does not hold in general if $x$ is not a positive real number or if $a,b$ are not both integers. A lot of people misunderstand this, and indeed there are many, many, many, many questions on this site that are rooted in this misunderstanding.

But with respect to the question in the OP: It is perfectly reasonable to argue that $i^4 = \left(i^2 \right)^2$, because even though $i$ is a complex number, the exponents are integers, so the basic notion of exponentiation as repeated multiplication is reliable.

• Can you add please the values for $i^9, i^{10}, i^{11}, i^{12}$ ? – ParaH2 Jan 5 '17 at 20:54
• @Hexacoordinate-C I can't tell if you're joking or not. – mweiss Jan 5 '17 at 20:57
• Note that this post uses the field properties of the complex numbers (namely that the nonzero complex numbers form a group under multiplication). See math.stackexchange.com/a/2085631 for the answer to the last part of the question about trying to apply logic that works for real numbers to complex numbers. – user21820 Jan 6 '17 at 3:48
• @user21820 Nice comment, +1 – 808GroundState Jan 6 '17 at 12:50
• You might also want to explain why the powers are cyclic because it wasn't instantly obvious why to me. – Travis Jan 6 '17 at 13:02

I'm surprised that none of the other answers pointed out the most important point in your question:

Obviously, this logic can be applied to any real numbers, but does it also apply to complex numbers?

This attitude is the right way to go. The logic you speak of is more precisely:

$x^{ab} = (x^a)^b$ for any real number $x$ and natural numbers $a,b$.

If you want the more general fact for integer exponents:

$x^{ab} = (x^a)^b$ for any real number $x \ne 0$ and integers $a,b$.

In fact it turns out that 'miraculously' we have an even more general fact for real exponents:

$x^{ab} = (x^a)^b$ for any real number $x > 0$ and reals $a,b$.

Notice that all these precise statements about real exponentiation show you clearly that you must know exactly what the objects are before you can apply any operations to them, not to say claim any properties about the resulting values.

For this reason it is actually an important question to ask whether there are corresponding rules for complex numbers.

Yes, but not as nice.

$x^{ab} = (x^a)^b$ for any complex number $x \ne 0$ and integers $a,b$. (*)

Here exponentiation is simply the result of starting from $1$ and repeatedly multiplying/dividing by $x$ where the number of times is specified by the exponent (multiplying for positive; dividing for negative). This fact holds in any structure that has invertible multiplication, including the field of rationals, the field of reals, and the field of complex numbers.

$x^{ab},x^a$ are well-defined since $x \ne 0$.

However, in general "$x^{ab} = (x^a)^b$" does not hold for complex $x$ even if $a,b$ are both rational. For instance (according to standard conventions):

$i = (-1)^{1/2} = (-1)^{(2 \times 1/4)} \ne ((-1)^2)^{1/4} = 1^{1/4} = 1$.

So it's excellent that you ask whether some new structure (complex numbers) have the same properties as some other structure (real numbers) instead of just blindly assuming it does.

The question was recently edited to ask for including an explanation of (*). Actually, there is nothing much to explain intuitively, since it boils down to the fact that an $ab$-fold repetition of an operation is the same as a $b$-fold repetition of an $a$-fold repetition of that operation. One can either stop there, but if one wants to ask why then one would need to fix a foundational system first, and in particular the rules concerning integers and induction/recursion. The below proof will use associativity of integer addition and multiplication, and distributivity of multiplication over addition for integers, which correspond to basic facts about repetition.

Suppose we have a field $S$ (such as the complex numbers) and an exponentiation operation that satisfies the following: $\def\lfrac#1#2{{\large\frac{#1}{#2}}}$

$x^0 = 1$ for every $x \in S$.

$x^{k+1} = x^k x$ for every $x \in S$ and integer $k$.

Note that any reasonable foundational system is capable of defining such an operation recursively (you need one direction for positive $k$ and another for negative $k$), and can easily prove by induction the following two theorems.

$x^{a+b} = x^a x^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.

Then $x^{a+0} = x^a = x^a x^0$.

Given any integer $b$ such that $x^{a+b} = x^a x^b$:

$x^{a+(b+1)} = x^{(a+b)+1} = x^{a+b} x = ( x^a x^b ) x = x^a ( x^b x ) = x^a x^{b+1}$.

$x^{a+(b-1)} = x^{(a+b)-1} = x^{a+b} \div x = ( x^a x^b ) \div x = x^a ( x^b \div x ) = x^a x^{b-1}$.

Therefore by induction $x^{a+b} = x^a x^b$ for every integer $b$.

$x^{ab} = (x^a)^b$ for every nonzero $x \in S$ and integers $a,b$.

Take any nonzero $x \in S$ and integer $a$.

Then $x^{a \times 0} = x^0 = 1 = (x^a)^0$.

Given any integer $b$ such that $x^{ab} = (x^a)^b$:

$x^{a(b+1)} = x^{ab+a} = x^{ab} x^a = (x^a)^b (x^a) = (x^a)^{b+1}$.

$x^{a(b-1)} = x^{ab-a} = x^{ab} \div x^a = (x^a)^b \div (x^a) = (x^a)^{b-1}$.

Therefore by induction $x^{ab} = (x^a)^b$ for every integer $b$.

Notice that we did not use commutativity here, which in fact shows that the argument holds in any division ring. If you restrict the exponents to natural numbers, then it clearly holds in any group when "nonzero" is deleted.

Finally, there are some nice properties that arise from the above properties such as:

$i^{k+4} = i^k i^4 = i^k (i^2)^2 = i^k (-1)^2 = i^k$ for any integer $k$.

In short, powers of $i$ (a square-root of $-1$ in the complex field) are cyclic.

• You give a very important message. All too often, a “rule” is taught as something universally true, without any mention of the domain of applicability. – Lubin Jan 6 '17 at 4:06
• @Lubin: Thank you! This is one of the things that makes teaching type checking (borrowed from CS) an extremely attractive notion to me, since it forces students to know what they are doing rather than mimicking their teacher and hoping what they write on paper gets the marks. =) – user21820 Jan 6 '17 at 4:12
• The answer is misleading: roots are the inverse of exponents so $i^4 = 1 = 1^{\frac{1}{4}}$. The problem is that the root operation is not one-to-one and this answer only took the principal root of $i^{\frac{1}{4}}$. It has 4 roots: $1, -1, i, -i$ – brent.payne Jan 7 '17 at 20:39
• @brent.payne: It is not misleading. That is the usual definition of the complex exponentiation operation. The point in my post is very clear; some properties of real exponentiation do not carry over to complex exponentiation. If you want a relation instead of a function, you cannot use the exponentiation notation. Also , even if you attempt to define exponentiation to be on sets of complex numbers, this same property will still fail; $\{1\} = \{1\}^{2 \times \frac12} \ne (\{1\}^2)^{\frac12} = \{-1,1\}$. You cannot anyhow choose a root you like. – user21820 Jan 8 '17 at 3:34
• The answer is very good except for the $i = (−1) ^ {1/2}$ and $1 ^ {1/4} = 1$ parts. This is not a standard convention. You own comment above says "You cannot anyhow choose a root you like.", yet you have chosen roots twice. – ypercubeᵀᴹ Jan 13 '17 at 9:05

Geometrically, multiplication by $i$ does the following to a complex number:

• Scales length by a factor of $1$
• Rotates 90 degrees

If you rotate by 90 degrees four times in the same direction, where do you end up?

• You should clarify this with a complex number plotted on the complex plane. – noobermin Jan 5 '17 at 23:47
• Obviously depending on where you start the rotation, so no one knows what $i^4$ is actually – Carlos - the Mongoose - Danger Jan 13 '17 at 4:56
• @BradPitt Neal is thinking of multiplication by a number as a function. Multiplication by $i^4$ is the identity function- scale by 1 and rotate by 0. – Daniel Moskovich Sep 12 '17 at 7:15
• @BradPitt The answer is, "The same way you were pointing when you started." In other words, nothing changed, so you multiplied by $1$. :) – Neal Sep 12 '17 at 10:26

Seeing as the current best answer that I understand doesn't contain proof at my level (and likely the level of the people this question would help), I'm going to answer.

The solutions of $i^n$ are repeating in a simple pattern. The pattern goes as follows

$$i^0 = 1$$ $$i^1 = i$$ $$i^2 = i\cdot i = -1$$ $$i^3 = (i\cdot i)\cdot i = -i$$ $$i^4 = i\cdot i\cdot i\cdot i = (i\cdot i)(i\cdot i) = (-1)^2 = 1$$

This pattern is repeated infinitely.

edit: As tomazs pointed out, this only works because multiplication by pure and simplified complex numbers is associative.

$$i\cdot i\cdot i\cdot i = (i\cdot i)(i\cdot i) = (i\cdot i\cdot i)\cdot i = ((i\cdot i)\cdot i)\cdot i$$ etc.

• Good answer. Just be exact with parenthesis. In the last step it is $(-1)^2=1$. Since otherwise one reads it as $-(1)^2=-1$ and this can lead to confusion. – Fritz Jan 5 '17 at 21:39
• @MarvinF, thanks for pointing that out, didn't even think about it. – Travis Jan 5 '17 at 21:42
• I think you were setting up a nice pattern with $i^2$ and $i^3$: To get to the next power of $i$, multiply the previous one by $i$. But you didn't continue that pattern for $i^4$, and compute it as $i^3 \cdot i$. There's nothing wrong with what you have, but "thematically," you might consider the "multiply by another copy of $i$" approach. – pjs36 Jan 6 '17 at 0:41
• @pjs36 I had that originally, but it looked even messier than my current proof of $i^4$ – Travis Jan 6 '17 at 0:56
• A fair point, and a rather sophisticated one :) I just wanted to mention it – pjs36 Jan 6 '17 at 0:58

Alternatively:

$$i^4=i^{2+2}=i^2i^2=(-1)(-1)=1$$

Raising to positive integer power is the same as repeated multiplication, thus you don't even have to think whether $i^4=(i^2)^2$ is true, just expand the power:

$$i^4=i\cdot i\cdot i\cdot i=(i\cdot i)(i\cdot i)=(-1)(-1)=1.$$

The second equality works due to associativity of multiplication for complex numbers.

• I prefer @Tomasz answer, that explicitly invokes associativity. Your parenthesing isn't justified. – Yves Daoust Jan 7 '17 at 15:02
• @YvesDaoust you're right. I was thinking of mentioning it, but for some reason decided to not do when I was writing the answer. Anyway, tomasz's answer is way more complete than mine, and it was written later, so I couldn't have had it as an example :). – Ruslan Jan 7 '17 at 15:18
1. $$i^4=(i^2)^2=(-1)^2=1$$
2. $$i^4=\left(|i|e^{\arg(i)i}\right)^4=\left(e^{\frac{\pi i}{2}}\right)^4=e^{\frac{4\pi i}{2}}=e^{2\pi i}=1$$
• Pretty sure that if OP is at the level of wondering whether $i^4 = 1$, they won't be familiar with the complex exponential. I'm quite often wrong though, so who knows... – pjs36 Jan 5 '17 at 20:54
• Proofing this with exponentiation looks like a cyclic proof to me – WorldSEnder Jan 5 '17 at 21:43

The property of multiplication of real numbers which you used and referred to, but did not precisely quote, is associativity (i.e. $z_1\cdot (z_2\cdot z_3)=(z_1\cdot z_2)\cdot z_3)$). Multiplication of complex numbers is associative, so you have: $$i^4=i\cdot(i\cdot(i\cdot i))=(i\cdot i)\cdot(i\cdot i)=(-1)\cdot (-1)=1.$$ (Note that if multiplication was not associative, an expression like $i^4$ would not immediately make sense: it is not at all obvious if it means $i\cdot(i\cdot(i\cdot i))$ or $((i\cdot i)\cdot i)\cdot i$. There are expansions of complex numbers with non-associative multiplication for which the rule you are using here would not apply.)

You can also verify it directly using the definition of multiplication of complex numbers: \begin{align} i\cdot(i\cdot(i\cdot i))&=(0+1i)\cdot ((0+1i)\cdot(-1+0i))\\ &=(0+1i)\cdot(0+(-1)i)\\ &=-(-1)+0i\\ &=1 \end{align}

Checking associativity in general is a bit more troublesome (in fact, it is almost always tiresome to check directly), but still workable. Recall that the definition of multiplication in complex numbers is $$(a_1+b_1i)\cdot(a_2+b_2i)=(a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2).$$ Now you can just compute directly \begin{multline*} (a_1+b_1i)\cdot \left( (a_2+b_2i)\cdot (a_3+b_3i)\right)=(a_1+b_1i)\cdot((a_2a_3-b_2b_3)+i(a_2b_3+b_2a_3))=\\ =(a_1a_2a_3-a_1b_2b_3-(b_1a_2b_3+b_1b_2a_3))+i(a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3). \end{multline*} On the other hand, \begin{multline*} ((a_1+b_1i)\cdot (a_2+b_2i))\cdot (a_3+b_3i)=((a_1a_2-b_1b_2)+i(a_1b_2+b_1a_2))\cdot(a_3+b_3i)=\\ =(a_1a_2a_3-b_1b_2a_3-(a_1b_2b_3+b_1a_2b_3))+i(a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2b_3).\end{multline*} Finally, $$a_1a_2a_3-a_1b_2b_3-(b_1a_2b_3+b_1b_2a_3)=a_1a_2a_3-b_1b_2a_3-(a_1b_2b_3+b_1a_2a_3),$$ and $$a_1a_2b_3+a_1b_2a_3+b_1a_2a_3-b_1b_2b_3=a_1a_2b_3-b_1b_2b_3+a_1b_2a_3+b_1a_2a_3,$$ so, since two complex numbers are equal when their real and complex parts are equal, the multiplication is associative. (This argument can be made to be a little bit shorter if you apply commutativity of multiplication of complex numbers.)

Yes you are right. Rotation by 90 degree in complex plane and multiplication by $i$ are very much same. If you raise $i$ to 4,8,12th ... powers you get 1. Raising to 1, 5, 9 .. you get back $i$; raising to 2,6,10 powers gives you $-1$ and so on.

The property $$(x^{m})^n=x^{mn}\tag{1}$$ for $m$ and $n$ nonnegative integers holds whenever we are dealing with an associative operation with a neutral element $1$, like for complex multiplication. Indeed it holds for $n=1$, because $(x^{m})^0=1=x^{m0}$ by definition.

Recall that $x^n$ is defined recursively: $x^0=1$, $x^{k+1}=x^k\cdot x$.

Suppose property $(1)$ holds for $n$; then \begin{align} (x^{m})^{n+1}&= (x^m)^n\cdot x^m &&\text{definition of powers}\\ &=x^{mn}\cdot x^m &&\text{induction hypothesis}\\ &=x^{mn+m} && \text{rule of powers $(2)$}\\ &=x^{m(n+1)} && \text{property of integers} \end{align} The mentioned rule of powers is where associativity is used: $$x^{h+k}=x^h\cdot x^k\tag{2}$$

Again, this is true by definition when $k=1$. Suppose it holds for $k$; then \begin{align} x^{h+(k+1)}&=x^{(h+k)+1} && \text{property of integers}\\ &=x^{h+k}\cdot x &&\text{definition of powers}\\ &=\bigl(x^h\cdot x^k\bigr)\cdot x &&\text{induction hypothesis}\\ &=x^h\cdot\bigl(x^k\cdot x\bigr) &&\text{associativity}\\ &=x^h\cdot x^{k+1} &&\text{definition of powers} \end{align}

Therefore you are certainly allowed to say that $$i^4=(i^2)^2=(-1)^2=1$$

Light comes from the definition of the complex multiplication, which only involves real arithmetic:

$$(a+ib)(c+id):=ac-bd+(ad+bc)i.$$

Then

$$(0+1i)^2=(0+1i)(0+1i)=(\bar1+0i),\\ (0+1i)^3=(\bar1+0i)(0+1i)=(0+\bar1i),\\ (0+1i)^4=(0+\bar1i)(0+1i)=(1+0i).$$

• Downvoter, please justify. My answer is perfectly valid. – Yves Daoust Jan 13 '17 at 14:33
• I'm not the downvoter, so I've no idea about the reason for the downvote, but if you ask me, you definitely should not use "$\overline{1}$" for "$-1$" since in complex analysis "$\overline{z}$" denotes the conjugate of $z$. Also, I don't think your answer quite addresses the underlying question of what complex exponentiation satisfies. You can make it relevant by stating that in your answer the integer exponent is defined to mean repeated multiplication, and so you are simply evaluating the powers one by one. Makes sense? – user21820 Jan 13 '17 at 15:56
• $\bar 1$ makes a much nicer alignment than $-1$. Purely cosmetic. I didn't think of the confusion with conjugation, which I use to denote $^*$. The accepted answer just says that the powers are cyclically $i,-1,-i,1$ without any justification. IMO, it is the one that deserves a downvote. – Yves Daoust Jan 14 '17 at 7:48
• I also use "$z^*$" to denote the conjugate of $z$, but the overline notation is actually very common. As I said, I didn't downvote your answer, but I do think it doesn't address the underlying question. And I don't know who downvoted my answer either..... Not you right? – user21820 Jan 14 '17 at 8:52
• @Yves Well, since this is not a peer-reviewed journal, I suppose idiosyncrasies such as these are more important than clarity. – Lisa Jan 27 '17 at 22:37