Why does $i^{-4} = -1$? This arrises from the accepted answer of is $i^4$ equal to $-1$? In which the answerer states:

The powers of $i$ are cyclic ...

the pattern persists for negative exponents, as well. Specifically, $i^{-1}=\frac{1}{i}=-i$.

He then goes on to explain how $i$ and $-i$ are multiplicative inverts.

What does it mean for $i$ and $-i$ to be multiplicative inverses and how does that create the conslusion that $\frac{1}{i}=-i$?

  • $\begingroup$ You are mixing where you mean to write $i$ and when you mean to write $1$. Be especially careful whenever writing things that you are writing what you intend. E.g. in your title you write $i^{-4}=-\color{red}{1}$ but in your body you write $i^{-4}=-\color{red}{i}$. Also, you write $\frac{1}{i}=-\color{red}{1}$ which is incorrect. $\endgroup$
    – JMoravitz
    Commented Jan 19, 2017 at 3:00
  • $\begingroup$ @JMoravitz Fixed the typos $\endgroup$
    – Travis
    Commented Jan 19, 2017 at 3:02
  • $\begingroup$ If a and b are multiplicative inverses of each other then a times b is 1. $\endgroup$
    – randomgirl
    Commented Jan 19, 2017 at 3:03

4 Answers 4


Well, we have


This method can be applied to any negative power.

  • $\begingroup$ I guess I should've reworded the title after I wrote the last bit of the question, it was more about why Yves's proof works rather than alternative proofs, though +1 because this helped me. $\endgroup$
    – Travis
    Commented Jan 19, 2017 at 3:00

The "multiplicative inverse" of a number $z$ is basically a fancy way of writing $\frac{1}{z}$ or $z^{-1}$. That is, the number $z^{-1}$ such that $zz^{-1}=1$.

$i$ and $-i$ are multiplicative inverses because $i\times -i=1$. Divide by $i$ to get that $\frac{1}{i}=-i$ or $i^{-1}=-i$. Raise each side to the power $4$ to get $i^{-4}=(-i)^4=1$.


As for the question "What does it mean to be multiplicative inverses"

Let $(\Bbb F,+,\times)$ be a field (in our context, $\Bbb F$ can be the complex numbers and $+$ and $\times$ are the addition and multiplication we are used to)

We will use the symbol $0$ to denote the additive identity, i.e. $x+0=x=0+x$ for all $x$ and we will use the symbol $1$ to denote the multiplicative identity, i.e. $x\times 1 = x = 1\times x$ for all $x$

The additive inverse of $x$ is the unique element of $\Bbb F$ which we call $(-x)$ such that $x+(-x)=0=(-x)+x$

The multiplicative inverse of $x$ (if one exists) is the unique element of $\Bbb F$ which we call $x^{-1}$ such that $x\times x^{-1}=1=x^{-1}\times x$

In the context of finding $i^{-1}$, we know that $i^{-1}\times i = 1$ and we also know that $i^3\times i = i^4 = i^2\times i^2 = (-1)\times (-1)=1$ so both $i^{-1}$ and $i^3$ are multiplicative inverses of $i$. Due to the uniqueness of inverses, that implies that they must be equal.

As for why inverses are unique, suppose that $x$ has at least two inverses, say $\color{red}{x^{-1}}$ and $\color{blue}{x^{-1}}$.


$\color{blue}{x^{-1}}=\color{blue}{x^{-1}}\times (x \times \color{red}{x^{-1}})=(\color{blue}{x^{-1}}\times x) \times \color{red}{x^{-1}} = \color{red}{x^{-1}}$

which shows that they were actually the same to begin with.


I understand you to be asking for the intuitive answer, so I will provide a handle on the intuition.

Suppose you are standing facing east. You make a quarter turn to the left and you are facing north; another quarter turn and you're facing west; one more quarter turn and you're facing south; and one more quarter turn and you're facing east again, the way you started.

Now forget you ever heard of complex or imaginary numbers, and play along with me while I invent some notation.

From now on, we are going to use some funny names for things. We will call the direction east by the name $1$. In this context the symbol "$1$" has no meaning, it's just a name I made up for the direction east.

Likewise from now on I call north by the name $i$, west by the name $-1$, and south by the name $-i$. Again these are just funny names, they have no mathematical meaning.

Now when I mean to say "quarter turn to the left" instead I will say "multiply by $i$". And I'll use the standard mathematical convention that repeated multiplication is denoted by exponentiation.

So if I start facing east, I haven't made any turns at all, so I call this $i^0$. You can see that $i^1$ brings me to face north; $i^2$ is notation for two consecutive left quarter turns, so $i^2 = -1$. (Sounds familiar!) Then $i^3$ is notation for making three left quarter turns, which leaves me facing south; and finally $i^4$ brings me back to where I started, facing east.

Now if we call a quarter turn to the right "multiplying by $-i$", then $i^{-4}$ represents four right quarter turns starting from facing east, which brings me back to facing east.

Now it turns out to be mathematically true that multiplying any complex number by $i$ simply rotates that complex number counterclockwise through an angle of $\frac{\pi}{2}$ radians. You can prove this mathematically by via rotation matrices in 2 dimensions.

According to the introduction to that Wiki page, we can rotate a vector in Euclidean 2-space by multiplying the vector by the matrix

$$ \left[ {\begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} } \right] $$

In particular, a counterclockwise rotation through an angle of $\frac{\pi}{2}$ radians corresponds to multiplication by the matrix

$$ \left[ {\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} } \right] $$

You can convince yourself that in fact multiplication by this matrix has the same effect on complex numbers (interpreted as plane vectors) as does multiplication by $i$.

In other words, multiplying a complex number by $i$ is simply a gadget for keeping track of how many times we made a quarter turn left; and multiplying by $-i$ keeps track of how many times we made a quarter turn right.

From this point of view, the equation $i^2 = -1$ is simple and natural, and no longer some kind of mystery. Likewise, $i^4 = i^{-4} = 1$.

  • $\begingroup$ As soon as you labeled the directions I got it, thanks so much $\endgroup$
    – Travis
    Commented Jan 19, 2017 at 3:57
  • $\begingroup$ @Travis I'm so glad you found that helpful. $\endgroup$
    – user4894
    Commented Jan 19, 2017 at 4:34

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