# How do you simplify i^20?

Mar 20, 2018

$1$

#### Explanation:

We have:

${i}^{1} = i$

${i}^{2} = - 1$

${i}^{3} = - i$

${i}^{4} = 1$

${i}^{5} = i$

${i}^{6} = - 1$

${i}^{7} = - i$

${i}^{8} = 1$

You see a pattern? Let's try to define it mathematically.

Let's define a function. For any division function, our function will return its remainder. We call our function $\zeta \left(\frac{u}{v}\right)$, where $v \ne 0$. For example,

$\zeta \left(\frac{8}{2}\right) = 0$, or:

$\zeta \left(\frac{15}{4}\right) = 3$

Note: For these imaginary numbers sequences, $n \in \mathbb{N}$.

Anyways, we find $\zeta \left(\frac{n}{4}\right)$ for any ${i}^{n}$. If:

• $\zeta \left(\frac{n}{4}\right) = 0$, we know that ${i}^{n} = 1$

• $\zeta \left(\frac{n}{4}\right) = 1$, we know that ${i}^{n} = i$

• $\zeta \left(\frac{n}{4}\right) = 2$, we know that ${i}^{n} = - 1$

• $\zeta \left(\frac{n}{4}\right) = 3$, we know that ${i}^{n} = - i$

Lets confirm this. Take ${i}^{6}$. We know that ${i}^{6} = - 1$. According to our "remainder function,"

$\zeta \left(\frac{6}{4}\right) = 2$, and we confirm that ${i}^{6} = - 1$.

Let's tackle our problem. Here, $n = 20$, so we have:

$\zeta \left(\frac{20}{4}\right) = 0$. Looking at our key, we find that when $\zeta \left(\frac{n}{4}\right) = 0$, ${i}^{n} = 1$.

So ${i}^{20} = 1$.

Mar 20, 2018

$1$

#### Explanation:

Given $z \in \mathbb{C} = {i}^{n} : n \in \mathbb{N}$

$z = + 1 : \forall n {\mod}_{4} = 0$

$z = + i : \forall n {\mod}_{4} = 1$

$z = - 1 : \forall n {\mod}_{4} = 2$

$z = - i : \forall n {\mod}_{4} = 3$

In this case $n = 20$

$20 {\mod}_{4} = 0 \to z = 1$