Which of the following is equivalent to the expression? $i^{22}$

Question

Which of the following is equivalent to the expression? $i^{22}$

A.) $-1$ B.) $i$ C.) $1$ D.) $-i$

What is $i$? How could it have a exponent if it's an imaginary number?

Answer 1

To understand powers of $i$, remember that $i^2 = -1$. Thus, we have

\begin{align*} i^0 & = 1\\ i^1 & = i\\ i^2 & = -1\\ i^3 & = i^2i = -1i = -i\\ i^4 & = (i^2)^2 = (-1)^2 = 1 \end{align*}

Since $i^4 = 1$, $i^{4k} = (i^4)^k = 1^k = 1$ for any integer $k$. This observation allows you to write

$$i^{22} = i^{20 + 2} = i^{20}i^2 = (i^4)^5i^2 = 1^5i^2 = 1 \cdot i^2 = 1 \cdot -1 = -1$$

Answer 2

STRONG HINT:

$$i^{22}=(i^2)^{11}.$$

Because complex numbers can be multiplied together, a definition for $z^n$ is not troublesome for natural and indeed whole number $n$.

Answer 3

we have $i^2=-1$ thus $i^{22}=(-1)^{11}=-1$

Answer 4

we know $i^2=-1$ so we have that $$i^{22}=(i^2)^{11}=(-1)^{11}=-1$$

Answer 5

Pretty much all the answers so far have used the same reasoning, but I thought of it a little differently. For exponents of $i$ greater than $4$, I usually work out the value of the exponent modulo $4$. That is, $i^n = i^{(n\mod 4)}$.

The reasoning is simple. $i^4 = 1$, so there's a "periodicity" with "period" $4$ as far as exponents of $i$ go.

So the answer here is $i^{(22\mod 4)} = i^2 = -1$.

Just wanted to post this solution so the asker knows a nice general way to approach powers of $i$ from now on.

Answer 6

$$i^0=\color{red}{+1}$$ $$i^1=\color{magenta}{+i}$$ $$i^2=\color{green}{-1}$$ $$i^3=i^2.i=\color{blue}{-i}$$ $$i^4=i^3.i=\color{red}{+1}$$ $$i^5=i^4.i=\color{magenta}{+i}$$ $$i^6=i^4.i^2=\color{green}{-1}$$ $$i^7=i^4.i^3=\color{blue}{-i}$$ $$...$$ $$i^{4n+\color{red}0}=\color{red}{+1}$$ $$i^{4n+\color{magenta}1}=\color{magenta}{+i}$$ $$i^{4n+\color{green}2}=\color{green}{-1}$$ $$i^{4n+\color{blue}3}=\color{blue}{-i}$$