# How do you simplify i^101?

Nov 13, 2015

Factor out multiples of ${i}^{4}$, yields dividing by ${i}^{100}$ and obtaining $i$

#### Explanation:

Recall the rules of $i$. $i$ is defined such that:

$i = i$
${i}^{2} = - 1$
${i}^{3} = i \cdot {i}^{2} = - i$
${i}^{4} = {\left({i}^{2}\right)}^{2} = {\left(- 1\right)}^{2} = 1$

We also know that exponent rules function such that $\frac{{x}^{a}}{{x}^{b}} = {x}^{a - b}$. If we divide ${i}^{101}$ by some factor of ${i}^{101}$, such as ${i}^{x}$, where $0 \le x < 101$ and where $x$ is divisible by 4, we will yield ${i}^{101 - x}$

With this in mind, return to the initial problem. What is the largest power of $i$ we can factor out of ${i}^{101}$ (e.g. what is the highest $x$ divisible by 4 and less than 101? This should be obtainable by subtracting 1 from 101 as many times as it takes to yield a number divisible by 4. As it turns out, we reach that point at $x = 100$.

From here, the calculation is simple.

${i}^{101} = \frac{{i}^{101}}{1} = \frac{{i}^{101}}{{i}^{100}} = {i}^{101 - 100} = {i}^{1} = i$