7.1 Imaginary concepts -- playing with i

Let’s begin with a few very simple exercises designed to show how we apply the normal rules of algebra to this new, abnormal number.

Now let's try something a little more involved.

It is vital to remember that $i$ is not a variable, and this is not an algebraic generalization. You cannot plug $i=3$ into that equation and expect anything valid to come out. The equation ${\text{(3+2i)}}^2=5+\mathrm{12i}$ has been shown to be true for only one number: that number is $i$ , the square root of –1.

In the next example, we simplify a radical using exactly the same technique that we used in the unit on radicals , except that $a–1$ is thrown into the picture.

Example: Simplify $\sqrt{-\text{20}}$
$\sqrt{-\text{20}}$ = $\sqrt{(4)(5)(-1)}$	as always, factor out the perfect squares
= $\sqrt{4}$ $\sqrt{5}$ $\sqrt{-1}$	then split it, because $\sqrt{\text{ab}}$ = $\sqrt{a}$ $\sqrt{b}$
$=\mathrm{2i}$ $\sqrt{5}$	$\sqrt{4}$ =2, $\sqrt{-1}$ $=i$ , and $\sqrt{5}$ is just $\sqrt{5}$
Check
Is $\mathrm{2i}$ $\sqrt{5}$ really the square root of –20? If it is, then when we square it, we should get –20.
${\left({\mathrm{2i}}^{}\sqrt{5}\right)}^2=2^2{i}^2{5}^2=4*-1*5=-20$ It works!

The problem above has a very important consequence. We began by saying “You can’t take the square root of any negative number.” Then we defined $i$ as the square root of –1. But we see that, using $i$ , we can now take the square root of any negative number.