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Complex Numbers

Simplifying Powers of i

The powers of i are cyclic. Let's look at what happens when we raise i to increasing powers.

\begin{aligned}&i^{1}=i \\&i^{2}=-1 \\&i^{3}=i^{2} \cdot i=-1 \cdot i=-i \\&i^{4}=i^{3} \cdot i=-i \cdot i=-i^{2}=-(-1)=1 \\&i^{5}=i^{4} \cdot i=1 \cdot i=i\end{aligned}

We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by increasing powers, we will see a cycle of four. Let's examine the next four powers of i.

\begin{aligned}&i^{6}=i^{5} \cdot i=i \cdot i=i^{2}=-1 \\&i^{7}=i^{6} \cdot i=i^{2} \cdot i=i^{3}=-i \\&i^{8}=i^{7} \cdot i=i^{3} \cdot i=i^{4}=1 \\&i^{9}=i^{8} \cdot i=i^{4} \cdot i=i^{5}=i\end{aligned}

The cycle is repeated continuously: i,-1,-i, 1, every four powers.

Example 8

Simplifying Powers of i

Evaluate: i^{35}.

Solution

Since i^{4}=1, we can simplify the problem by factoring out as many factors of i^{4} as possible. To do so, first determine how many times 4 goes into 35: 35=4 \cdot 8+3.

i^{35}=i^{4 \cdot 8+3}=i^{4 \cdot 8} \cdot i^{3}=\left(i^{4}\right)^{8} \cdot i^{3}=1^{8} \cdot i^{3}=i^{3}=-i

Try It #7

Evaluate: i^{18}