Chapter 1.6, Problem 69PS

Solution

To make: A table that shows the powers of $i$ from $i^1$ to $i^8$ in the first row and the simplified forms of these powers in the second row. Also, to describe the pattern observed in the table and to verify that the pattern continues by evaluating the next four powers of $i$ .

Table is as follows.

  $\begin{array}{ccccccccc}\text{Powers}\text{of}i& i& i^2& i^3& i^4& i^5& i^6& i^7& i^8\\ \text{Simplified}\text{form}& i& -1& -i& 1& i& -1& -i& 1\end{array}$

It can be see that the pattern repeats every four powers of $i$ .

Given information:

Complex number is $i$ .

Calculation:

The first power of any number equals that number. Thus, $i^1=i$ . Since $i=\sqrt{-1},i^2=-1$ .

The third power $i^3$ can be written as a product of $i^2$ and $i$ .

  $\begin{array}{c}{i}^3=i^2\cdot i\\ =-1\cdot i\\ =-i\end{array}$

Similarly, evaluate the powers of $i$ from $i^4$ to $i^8$ .

  $\begin{array}{c}{i}^4={\left(i^2\right)}^2\\ ={(}^{-}\\ =1\end{array}$

Now,

  $\begin{array}{c}{i}^5=i^4\cdot i\\ =1\cdot i\\ =i\end{array}$

Now,

  $\begin{array}{c}{i}^6=i^4\cdot i^2\\ =1\cdot (-1)\\ =-1\end{array}$

Now,

  $\begin{array}{c}{i}^7=i^6\cdot i\\ =-1\cdot i\\ =-i\end{array}$

Now,

  $\begin{array}{c}{i}^8={\left(i^4\right)}^2\\ ={\left(1\right)}^2\\ =1\end{array}$

Organize the results in a table as shown.

  $\begin{array}{ccccccccc}\text{Powers}\text{of}i& i& i^2& i^3& i^4& i^5& i^6& i^7& i^8\\ \text{Simplified}\text{form}& i& -1& -i& 1& i& -1& -i& 1\end{array}$

From the table, it can be seen that the pattern repeats every four powers of $i$ .