# What are imaginary numbers?

We’ve always been told we can’t take the square root of a negative number, since the square of a negative number is a positive number. But remember, in maths nothing is impossible, not even dividing by 0!

Imagine you want to take the square root of -1;

$\sqrt{-1}$

or the square root of -4;

$\sqrt{-4}$

or any other negative number. These are made up of -1 and their absolute value:

$-4=(-1)|4|$

This means that we can consider -1 the negative unit. So, the square root of -4 is

$\sqrt{-4}=\sqrt{(-1)\cdot&space;4}=\sqrt{-1}\cdot&space;\sqrt{4}=2\sqrt{-1}$

And, more in general:

$\sqrt{n}=\sqrt{(-1)\cdot&space;|n|}=\sqrt{-1}\cdot&space;\sqrt{|n|}$

with n < 0.

But what is the square root of -1? It is called i, known as the imaginary unit. Therefore:

$\sqrt{n}=\sqrt{(-1)\cdot&space;|n|}=\sqrt{-1}\cdot&space;\sqrt{|n|}=i\sqrt{|n|}$

And in the case of -4 we have

$\sqrt{-4}=\sqrt{(-1)\cdot&space;|-4|}=\sqrt{-1}\cdot&space;\sqrt{4}=2i$

Another example:

$\sqrt{-8}=\sqrt{(-1)\cdot&space;|-8|}=\sqrt{-1}\cdot&space;\sqrt{8}=2i\sqrt{2}$

This is cool, isn’t it?

2i, 3i, -8i and so on are imaginary numbers, made up of the imaginary unit and the real numbers.

If the square root of -1 is i, i² = -1, and so i times i is equal to -1 as well. But what is i³?

$i^3=i\cdot&space;i\cdot&space;i=(i\cdot&space;i)\cdot&space;i=(-1)\cdot&space;i=-i$ $i^4=\left&space;(&space;i^{2}&space;\right&space;)^2=(-1)^2=1$

And you can try with other different powers. When it comes to addition and subtraction, you can think of i as a constant a:

$2i+i=3i$

the same way 2a + a = 3a. But

$2i+1\neq3i$

the same way 2a + 1 is not 3a.

You can factor out the i as well:

$5i-3i=i(5-3)=2i$

$3i\cdot&space;4=12i$ $3i\cdot&space;4i=-12$ $6i&space;:&space;2=3i$ $6i&space;:&space;2i=3$

When we divide two imaginary numbers, the i goes away as if it was an x.

When we add or subtract an imaginary number and a real number, we get a complex number. Examples of complex numbers are 4i + 3, -2 + 9i, -5i – 8, etc…, but I will talk about them more specifically in another post. In the meantime, subscribe and stay tuned for more! And if you have any doubt of questions don’t hesitate to leave a comment; I will be happy to help you!

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