# Powers of iota

Since \({i^2} = - 1\), we have:

\[\begin{align}&{i^3} = {i^2} \times i = - 1 \times i = - i\\&{i^4} = {\left( {{i^2}} \right)^2} = {\left( { - 1} \right)^2} = 1\end{align}\]

Now, we can calculate *i* raised to any integer power. For example,

\[\begin{align}&{i^{37}} = {i^{36}} \times i = {\left( {{i^4}} \right)^9} \times i = 1 \times i = i\\&{i^{99}} = {i^{96}} \times {i^3} = {\left( {{i^4}} \right)^{24}} \times {i^3} = 1 \times - i = - i\\&{i^{ - 1}} = \frac{1}{i} = \frac{i}{{{i^2}}} = \frac{i}{{ - 1}} = - i\\&{i^{ - 50}} = \frac{1}{{{i^{50}}}} = \frac{1}{{{i^{48}} \times {i^2}}} = \frac{1}{{{i^2}}} = - 1\end{align}\]

**Example 1:** Find the value of \({i^{4n + k}}\), where *n* and *k* are integers, and *k* is in the set {0, 1, 2, 3}.

**Solution:** We have:

\[{i^{4n + k}} = {i^{4n}} \times {i^k} = {\left( {{i^4}} \right)^n} \times {i^k} = 1 \times {i^k} = {i^k}\]

Thus, the value of \({i^{4n + k}}\) is the same as the value of \({i^k}\), which depends on the value of *k*:

\[\begin{align}&k = 0:\; {i^k} = \;1\\&k = 1:\; {i^k} = \;i\\&k = 2:\; {i^k} = - 1\\&k = 3: \;{i^k} = - i\end{align}\]

**Example 2:** (a) Find the value of \(\begin{align}{i^{500}} + {i^{501}} + {i^{502}} + {i^{503}}\end{align}\).

(b) Show that the sum of any four consecutive powers of iota is 0.

**Solution:** (a) We have:

\[\begin{align}&{i^{500}} = {\left( {{i^4}} \right)^{125}} = {1^{125}} = 1\\&{i^{501}} = {i^{500}} \times i = i\\&{i^{502}} = {i^{500}} \times {i^2} = {i^2} = - 1\\&{i^{503}} = {i^{500}} \times {i^3} = {i^3} = - i\end{align}\]

Clearly, the sum of these four terms is 0.

(b) We have:

\[\begin{align}&{i^n} + {i^{n + 1}} + {i^{n + 2}} + {i^{n + 3}}\\&= {i^n}\left( {1 + i + {i^2} + {i^3}} \right)\\&= {i^n}\left( {1 + i - 1 - i} \right) = 0\end{align}\]