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 $i^4=1$?
$\endgroup$ 12 Answers
$\begingroup$$i^3 = i i^2 = i(-1) = -i$
and
$i^4 = (i^2)^2 = (-1)^2 = 1$
$\endgroup$ 6 $\begingroup$The important thing to help you determine the value of $i^n$ for some integer $n$ is the exponent rule,
$$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$$
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