# i for Imaginary

Hello all!

I’d like to talk to all of you about imaginary numbers.  Can you think of an imaginary number?  It’s pretty hard huh?  Maybe if I gave you a number line you could point to where it might be?

Nope it’s not there!  That’s because all the numbers on the number line are real.  Integers, both positive and negative; rational numbers, which are numbers that can be expressed as a fraction; and irrational numbers, which are all the numbers in between the rational numbers, can all be found on the number line.  Therefore we must look elsewhere.  The number line is one-dimensional and we need to give ourselves another dimension to find imaginary numbers.

First, I want you to notice something in particular about how negative numbers interact with other numbers. If we multiply 3 by -1 it is easy to see that we get -3.  What might not be so obvious is that in the process we moved to a position on the number line that was 180 degrees counter clockwise from our original position.  Another way of stating this is that we went past the origin on our way to -3.  Multiplying two positive numbers together will never do this.

Note: When determining an angle we have to use three points: our starting point (red dot at 3), the origin (black dot), which we always use , and our ending point (blue dot at -3) to create an angle.
Now I want to show you what a two-dimensional “number line” might look like.  We still have the regular number line as the x-axis, but now we have an imaginary line as the y-axis.  “i” is the base unit of our new axis.  “i” stands for the $\sqrt{-1}$ and there can be multiples of “i” in either direction as shown below.

Notice that when we multiply 3 by one unit of “i” we get 3i.  If we make an angle from the starting point 3, go through the origin, and end at 3i we have created a 90 degree angle (shown in cyan below).  When we multiply by “i” again we get -3, which makes another 90 degree angle.  This process is cyclical.  We can multiply 3 by “i” raised to any power and we will end up with a number on the circle with a radius of 3.

This cyclic nature of i can be shown mathematically:

$i=\sqrt{-1}$

$i^2=\sqrt{-1}*\sqrt{-1}=-1$

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

$i^4=i^2*i^2=-1*-1=1$

$i^5=i^4*i^1=1*\sqrt{-1}=\sqrt{-1}$

Once we get to $i^5$ we realize that $i^5=i$ and cyclic pattern continues.

Now, there might be some who say that $\sqrt{-1}$ is not real.  I’m here to tell you that while it might not be a real number, it does exist.  We have to expand the set of numbers that we work with to include these new types of numbers.  This new set of numbers that encompasses both real and imaginary numbers is called complex numbers.

Complex numbers come in form of a+bi. This is a result of the fact that we are now working with a 2D plane and not a one dimensional line.  Any complex number can be represented on this plane by a real part “a” and an imaginary part “bi”.  These parts can be expressed as vectors along their respective axes and when added head to tail they produce a vector from the origin of the complex number they represent.

For example, the blue vector above is a complex number made up of 3 real units, represented by the red vector; and 3i imaginary units, represented by the green vector.  We would list the point as 3+3i.  This is different than a normal x-y coordinate plane because a point in a normal x-y plane represents two real numbers.  Here we’re only graphing one complex number which has two parts.  As you can imagine this gets complicated extremely fast if you want to graph functions of complex numbers; however, this is only an introduction so we will not go into more detail on how that works.

Let’s take a closer look at the generic complex number: a+bi.  Notice that if $b=0$ the complex number becomes real and is represented by only the red vector.  Put another way, real numbers can be considered a subset of complex numbers.  If $a=0$ the complex number is said to be purely imaginary!

<- A plant with square roots.

So what can we do with i ?

We can use it to get solutions to problems that end up taking the square root of a negative number.  For instance,

$x^2=-49$

$\sqrt{x^2}=\sqrt{-49}$

$x=\pm\sqrt{-49}$

$x=\pm\sqrt{49}*\sqrt{-1}$

$x=\pm 7i$

Here is a classic example of where you might run into imaginary roots.  Take a look at the function below with the equation: $y=x^2+3$.  If you were to try and solve for the roots of the equation, you would find that you end up with $x= \sqrt{-3}$.  Using the set of all real numbers we would be unable to determine the roots of this function.  Indeed, the function never goes through the x-axis.  However, by increasing our acceptable set of numbers to include imaginary ones, we are able to say that the roots are $\pm 3i$.

One issue that I see students struggle with is getting an imaginary term out of the denominator:

$\dfrac{3x}{4-i}$

All we have to do is multiply by 1!  To be more specific, we have to multiply by 1 in a disguised form: $\dfrac{4+i}{4+i}$

It’s the same as multiplying by $\dfrac{2}{2}$ or $\dfrac{9.33}{9.33}$.

By using the conjugate of the denominator we eliminate a middle term when we FOIL.

$\dfrac{(3x)*(4+i)}{(4-i)*(4+i)}$

$\dfrac{12x+3xi}{16+4i-4i-i^2}$

$\dfrac{12x+3xi}{16-i^2}$

$\dfrac{12x+3xi}{16-(-1)}$

$\dfrac{12x+3xi}{17}$

If you enjoyed reading this post or thought that it helped you understand the concept better, please leave a comment!  If you would like me to cover another topic you can request it here.

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