Complex Numbers: What Is i?

$i$ : The imaginary number

We'll talk about $i$, which is also called the imaginary number.

Now, suppose we want to find a real number $x$ such that $x^2 = 4$. Sounds easy right? In fact, there are two such real numbers.

Ok, that was easy. Now, try solving $x^2 = -1$.

Hmmmm, we got a little problem.

NO! So, we have run into quite a problem here. There seems to be no solution in real numbers.

This is where $i$ comes in. $i$ is defined as the number such that $i^2 = -1$, or informally written as $\sqrt{-1}$, the "square root" of $-1$. (which is actually not defined)

What is $i$ used for. For one thing, it plays a vital role for the set of complex numbers, which has the important property of algebraic closure.

However, $i$ is not real, it is imaginary. That's why it is called the imaginary number.

$Imaginary$ $Numbers!!!$

We've all been told in our Algebra 1 class that we cannot take the square root of an negative number.

However, this is not exactly true. You can take the square root of a negative number, but it involves using a new number to do it. This new number is called $i,$ standing for "imaginary."

The imaginary is defined to be:

$i=$$\sqrt{-1}$

Then:

$i^2$$=$$-1$

The Discovery of $i$

When mathematicians were solving quadratic equations, they had an interesting dilemma. If we try to solve the equation $x^2 + 1 = 0$, we get that the solutions are $x = \pm\sqrt{-1}$! However, mathematicians knew that no "real" number could have a square of $-1$, so they invented another number system just for this.

Mathematicians defined the imaginary unit as $i$. It is a number such that $i^2 = -1$. Then, mathematicians could suddenly solve the quadratic! The roots were just $x = \pm i$. They could also solve things like $x^2 + 4 = 0$; the solutions to that were just $\pm 2i$.

However, when these numbers were first introduced, many mathematicians were skeptical. They called these numbers "imaginary" and not like the "real" numbers that were used then. When these numbers became more widely accepted, the name stuck. Real numbers are the set of all rational and irrantional numbers, while imaginary numbers are numbers of the form $ai$ for some real number $a$ (some sources call these pure imaginary).

Now, let's have a short quiz:

That's all for this explanation. Thanks for reading my explanation about $i$ and imaginary numbers!

You know how to take square roots of numbers: $$\sqrt{9} = 3 \\ \sqrt{16} = 4$$ You also know that sometimes the resulting numbers are irrational.

However, let's try taking the square root of a negative number, say $-9$. Basically, we need to find a number $x$ such that $$\sqrt{-9} = x$$ Squaring the equation gives us $$-9 = x^2$$ However, this goes against our intuition! For any positive number $x$, $x^2$ will be positive, and when $x$ is negative, $x^2$ is also positive! (If $x = 0$ then $x^2$ is obviously $0$.) Therefore, we might be tempted to say that this has no solution for $x$!

Well, that would be true if $x$ is a real number. However, for now, let's try something: $$\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3\sqrt{-1}$$ Look! This form looks better! If only we could do something about that $\sqrt{-1}$...

Actually, that's about it. We can't simplify this anymore, so let's define a constant $i = \sqrt{-1}$. Our solution is now $3i$.

This extension using a new constant $i$ is known as the field of imaginary numbers. Imaginary numbers are monomials containing this constant $i$ e.g. $3i$, $5i$, $-8i$.

Complex numbers are numbers with two parts: a real part and an imaginary part:

$a+bi$ where $a$ and $b$ are real.

It will be discussed more in the next section!

That's why, mathematicians decided to use $i$ to represent the value of $\sqrt{-1}$, since it is imaginary and it doesn't exist.

Now we often call that a complex number, which is in the form $a+bi$