# 9.1 Playing with i

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This module contains some example problems involving the manipulation i, the imaginary number.

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.

 A few very simple examples of expressions involving $i$ Simplify: $i•5$ Answer: $\mathrm{5i}$ Simplify: $i+\mathrm{5i}$ Answer: $\mathrm{6i}$ (Add anything to 5 of itself, and you get 6 of it. Or, you can think of this as “pulling out” an $i$ as follows: $i+\mathrm{5i}=i\left(1+5\right)=\mathrm{6i}$ ) Simplify: $\mathrm{2i}+3$ Answer: You can't simplify it.

Now let's try something a little more involved.

Example: Simplify the expression (3+2i)2
${\left(3+\mathrm{2i}\right)}^{2}={3}^{2}+2\left(3\right)\left(\mathrm{2i}\right)+{\left(\mathrm{2i}\right)}^{2}$ because $\left(x+a\right)2={x}^{2}+2\mathrm{ax}+{a}^{2}$ as always
$=9+\mathrm{12i}–4$ ${\text{(2i)}}^{2}=\text{(2i)}\text{(2i)}=\text{(2)}\text{(2)}\text{(i)}\text{(i)}={\mathrm{4i}}^{2}=–4$
$=5+\mathrm{12i}$ we can combine the 9 and –4, but not the $\mathrm{12i}$ .

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 $\sqrt{-\text{20}}$
$\sqrt{-\text{20}}$ = $\sqrt{\left(4\right)\left(5\right)\left(-1\right)}$ 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.

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