# Imaginary numbers

Quadratic equations as $x^2 = -1$ don't have any real number solution. The solution exists in the complex number system.

Backbone of which is the imaginary unit $i$. For which following is true:

• $i = \sqrt{-1}$
• $i^2 = -1$

By taking multiples of $i$, I can create infinitely many pure imaginary numbers, e.g. $3i$, $i\sqrt{5}$, $-12i$.

By squaring an imaginary number I can explore how those relate to the real numbers: $$(3i)^2 = 3^2i^2 = 9i^2$$

Now I can apply the fact that $i^2 = -1$:

$$9i^2 = 9(-1) = -9$$

Meaning $3i$ is square root of $-9$

## Simplifying pure imaginary numbers

As proved $\sqrt{-9}$ simplifies to $3i$ and following though process can be applied: The square root of $-9$ is an imaginary number. The square root of $9$ is $3$, so the square root of negative $9$ is $3$ imaginary units, or $3i$

More formally: $$\text{For}~ a > 0, \sqrt{-a} = i\sqrt{a}$$

## Powers of $i$

Basic power line can be devised: \begin{aligned} i^0 &= 1 \\ i^1 &= i \\ i^2 &= -1 \\ i^3 &= i^2 \times i = (-1)(i) = -i \\ i^4 &= i^3 \times i = (-i)(i) = (-1)(i)(i) = (-1)(-1) = 1 \\ i^5 &= i^4 \times i = 1 \times i = i \\ i^6 &= i^5 \times i = -1 \\ i^7 &= i^6 \times i = -i \\ i^8 &= i^7 \times i = 1 \\ \end{aligned}

Results can be summarised in following table: