# How do you simplify i^109?

Dec 12, 2015

${i}^{100} = 1$

#### Explanation:

We will be using the following:

• ${\left({x}^{a}\right)}^{b} = {x}^{a b}$

• ${i}^{2} = - 1$

Because ${i}^{2} = - 1$, we know that ${i}^{4} = {\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$

Using this, we can take $i$ to a large integer power and quickly evaluate it by "pulling out" multiples of $4$ from the exponent, like so:

${i}^{100} = {i}^{4} \cdot 25 = {\left({i}^{4}\right)}^{25} = {1}^{25} = 1$

While the above suffices for the given problem, it is also applicable to integers which are much larger, or not multiples of $4$. For example, looking at ${i}^{528533}$

We can write

$528533 = 528500 + 32 + 1$
$= 100 \cdot 5285 + 4 \cdot 8 + 1$
$= 4 \cdot \left(25 \cdot 5288\right) + 4 \cdot 8 + 1$
$= 4 \left(25 \cdot 5288 + 8\right) + 1$

So, applying this to our problem:

${i}^{528533} = {i}^{1 + 4 \left(25 \cdot 5288\right)}$
$= {i}^{1} \cdot {\left({i}^{4}\right)}^{25 \cdot 5288}$
$= i \cdot {1}^{25 \cdot 5288}$
$= i \cdot 1$
$= i$