# What is the answer?

## Imaginary numbers:

Nov 14, 2017

${i}^{0} = 1$
${i}^{1} = i$
${i}^{2} = - 1$
${i}^{3} = - i$
${i}^{4} = 1$

#### Explanation:

Generally if you want to get the solution to ${i}^{x}$ where $x$ is any non-negative integer, you need to remember the following:

${i}^{0} = 1$
${i}^{1} = i$
${i}^{2} = - 1$
${i}^{3} = - i$
${i}^{4} = 1$

For all non-negative integers, if it is an exponent of $i$ then you simply have to divide $x$ by 4, 3, or 2.

If $x$ is divisible by 4 then ${i}^{x} = 1$
If $x$ is divisible by 3 then ${i}^{x} = - i$
If $x$ is divisible by 2 then ${i}^{x} = - 1$
If $x$ is not divisible by any of the above numbers then ${i}^{x} = i$
If $x$ is 0 then ${i}^{x} = 1$ since anything raised to 0 is equal to 1
Note that for this you must choose the highest number among 4, 3, and 2 that $x$ is divisible with.

Similarly,
${i}^{2} = i \cdot i = - 1$, so it also goes that:
${i}^{3} = i \cdot i \cdot i = {i}^{2} \cdot i = - 1 \cdot i = - i$
${i}^{4} = i \cdot i \cdot i \cdot i = {i}^{2} \cdot {i}^{2} = - 1 \cdot - 1 = 1$

Nov 14, 2017

${i}^{0} = 1$
${i}^{1} = i = \sqrt{- 1}$
${i}^{2} = - 1$
${i}^{3} = - i = - \sqrt{- 1}$
${i}^{4} = 1$

#### Explanation:

$i = \sqrt{- 1}$

${i}^{0} = \left({x}^{0} = 1 | x \in C\right)$ ... any number (x) power 0 is 1
${i}^{1} = i$ ... any number (x) power 1 is the number (x)

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

${i}^{3} = i \cdot {i}^{2} = \sqrt{- 1} \cdot 1 = \sqrt{- 1} = i$

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

note: for any ${i}^{a} = b$, one can write
$\left({i}^{2 n}\right) \left({i}^{a - 2 n}\right) = b$
when
$2 n$ is Even
AND
$2 n - a = 1 \mathmr{and} 2 n - a = 3$

for exmple:
i^35=?

We will take $\left(2 n = 32 , 2 n - a = 3\right)$

${i}^{35} = {i}^{32} \cdot {i}^{3} = \left({\left({i}^{4}\right)}^{8}\right) \cdot {i}^{3} = {\left(1\right)}^{8} \cdot {i}^{3} = 1 \cdot {i}^{3} = {i}^{3} = i$

end note