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# System prepended metadata

title: 'Imaginary Numbers: The Tip of the Iceberg'

---

# Imaginary Numbers: The Tip of the Iceberg

You might be here because you heard someone say numbers were **imaginary** and you took that literally. Well, there are indeed, although poorly named, **imaginary numbers.**

But what are these seemingly useless numbers for? What **are** these numbers?

Let's start with an example:
$$
x^2 + 4 = 0
$$

If you're still in Algebra I and would like to instill dreadful fear into the eyes of your unsuspecting teacher, ask them how to solve this with the methods they've taught so far.

Let's try to solve it as you would without any fancy methods.

$$
\begin{equation}
\begin{split} x^2 + 4 &= 0\\
x^2 &= -4 \\
\sqrt{x^2} &= \sqrt{-4} 
\end{split}
\end{equation}
$$

Uh... now what?
You may think: "isn't it -2?"

You'd be correct if it was:
$$
\sqrt{x^2} = -\sqrt{4}
$$


However, a simple check proves otherwise: $-2^2 \neq 4$. 
(Sidenote: Keep in mind that by the Order of Operations, the $2^2$ would be calculated first, so the answer is $-4$)

Regular numbers won't work, since any real number squared is positive.

### Multiplicative Sign Chart:
|  a  |  b  |  =  |
| --- | --- | --- |
| $+$ | $+$ | $+$ |
| $-$ | $-$ | $+$ |
| $+$ | $-$ | $-$ |
| $-$ | $+$ | $-$ |

If you square a number, the signs used to multiply will be the same, therefore, you cannot have a square that is negative with only real numbers.



To solve this, we'll need our beloved **imaginary numbers**.

Imaginary numbers are expressed in the imaginary unit, $i$. 
While it may appear confusing, *i* is not a variable, but a unit.
It is anagolous to the number $1$ in the real number system.
It is a factor of all imaginary numbers, just as 1 is.
$i$ is always conventionally after its "number", like so:

- $-2i$
- $7i$
- $6xi$

The imaginary unit has one defining property:

$$
i^2 = -1
$$

This, is what allows negative squares.

The rules of arithmetic apply as normal (for the most part).

$$
\begin{equation}
\begin{split} 
3i + 2i &= 5i \\
-7i + 5i &= 2i \\
\newline
3i \cdot 2i &= 6i \\
7i \cdot -3i &= -2i \\
\newline
\frac{4i + 8i}{2} &= 2i + 4i
\newline
\newline
i^1 &= i \\
i^2 &= -1 \\
i^3 &= -1 \cdot i \\
&= -i \\
i^4 &= i^2 \cdot i^2 \\
&= -1 \cdot -1 \\
&= 1 \\
i^5 &=1 \cdot i \\
&= i
\end{split}
\end{equation}
$$
    
The powers seem a little odd, but they repeat in that same cycle infinitely, which may take a little getting used to. Keep in mind that $i = \sqrt{-1}$.

With this, you are now armed with the knowledge to solve those nasty negative square roots!

Let's use our newfound knowledge on the example problem:

$$
\begin{equation}
\begin{split} x^2 + 4 &= 0\\
x^2 &= -4 \\
\sqrt{x^2} &= \sqrt{-4} \\
x &= \pm2i 
\end{split}
\end{equation}
$$

If you're confused, we use the $\pm$ notation as shorthand for both the positive and negative version of that expression.

If you square root a number, you **must** have a $\pm$ for its root, as the negative version of the number also results in the same square.

$$
\begin{equation}
\begin{split} 
5 \cdot 5 &= 25 \\
-5 \cdot -5 &= 25 \\
\end{split}
\end{equation}
$$

That's a very brisk introduction to imaginary numbers. While you may be able to use them well enough for your currently simple tasks, we haven't even seen half of what's to come. What happens if you use real numbers **and** imaginary numbers... together? The, again, poorly named: **complex numbers.**

That's enough for today though.
Thank you for reading, and have a good day!