# How do you simplify i^41?

Jul 7, 2018

See explanation.

#### Explanation:

To simplify this expression first let's calculate some low powes of $i$:

• ${i}^{2} = - 1$

• ${i}^{3} = - i$

• ${i}^{4} = 1$

From this calculations we can write that:

${i}^{41} = {i}^{40} \cdot i = {\left({i}^{4}\right)}^{10} \cdot i = {1}^{10} \cdot i = i$

Jul 7, 2018

$i$

#### Explanation:

We know that

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

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

${i}^{4} = 1$

This may be a less intuitive way of going about this, but let's see:

The imaginary unit follows a pattern. From ${i}^{1}$ to ${i}^{4}$, it goes

$i , - 1 , - i , 1$

Every time the exponent increases by $4$, we start the pattern over. This means that when our power of $i$ is

$5 , 9 , 13 , 17 , 21 , 25 , 29 , 33 , 37 , \textcolor{b l u e}{41}$

We will be equal to $i$. Now we see that ${i}^{41} = i$

Hope this helps!