Complex Numbers

This summary of algebraic operations on complex numbers will prepare you for solving quadratic equations with no solutions and the related implications for graphing quadratic and polynomial functions.

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}