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I know $i=\sqrt{ -1}$; what I don't get is the results you get from raising $i$ to an exponent:

$i^1 = i$ makes sense since anything to the first is itself.

$i^2 = -1$ also makes sense since squaring a square root is the same as removing the radical.

Why does $i^3=-i$ and $i^4=1$?

Isaiah
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2 Answers2

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$i^3 = i i^2 = i(-1) = -i$

and

$i^4 = (i^2)^2 = (-1)^2 = 1$

SometimesBlind
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  • That's a reasonable way to look at it. – Isaiah Mar 05 '15 at 23:58
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    This is wholly correct, but it may be worth saying here that the identity $x^{ab} = (x^a)^b$ used here applies because $a, b \in \mathbb{Z}$, which does not hold in general for complex $x, a, b \in \mathbb{C}$. In fact it can generally depend on a choice of "branch cut" of the involved functions $w \mapsto w^c$ where $c \not\in \mathbb{Z}$. – Travis Willse Mar 06 '15 at 00:04
  • @Travis is that also true for $a,\ b \in \Bbb{R}$, or just $\Bbb{Z}$? – user3002473 Mar 06 '15 at 03:54
  • @user3002473 It is not even generally true for $a \in \mathbb{Z}, b \in \mathbb{Q}$: If one takes $a = 2$ and $b = \frac{1}{2}$ and takes chooses the branch cut of $w \mapsto w^{1/2}$ (that is, the square root function) so that the image takes values with argument in $(-\frac{\pi}{2}, \frac{\pi}{2})$, then for $z$ in the right half-plane we have $(z^2)^{1/2} = z$ but for $z$ in the left half-plane we have $(z^2)^{1/2} = -z$. – Travis Willse Mar 06 '15 at 06:09
  • @Travis I think this is only an issue if you see rooting as a function.because you “throw away“ one of two possible outcomes. If you see it just as the inverse operator of squaring, you are fine again. – SometimesBlind Mar 06 '15 at 13:00
  • @SometimesBlind It certainly is a function. The point is that on the complex numbers (indeed, even just on $\mathbb{R}$), squaring is not invertible in the first place, as we always have $z^2 = (-z)^2$. – Travis Willse Mar 06 '15 at 14:59
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The important thing to help you determine the value of $i^n$ for some integer $n$ is the exponent rule,

$$a^{x+y}=a^xa^y$$

Hence for a quantity like $i^3$, you can rewrite $i^3 = i^{2+1}=i^2i^1$. You already know $i^2=-1$ and $i^1=i$ so it follows that $$i^2i^1 = (-1)(i) = -i = i^3$$ Similarly, $i^4 =i^{2+2} = i^2i^2$.

It might also be helpful to keep in mind that $$a^{xy} = (a^x)^y$$ That will help you figure out the value of something like $i^{28}$, since you could rewrite $$i^{28} = i^{4\cdot 7} = (i^4)^7 = (1)^7 = 1$$

graydad
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