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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: | $i\u20225$ |
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: $i+\mathrm{5i}=i(1+5)=\mathrm{6i}$ ) |
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 | |
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${(3+\mathrm{2i})}^{2}={3}^{2}+2\left(3\right)\left(\mathrm{2i}\right)+{\left(\mathrm{2i}\right)}^{2}$ | because $(x+a)2={x}^{2}+2\mathrm{ax}+{a}^{2}$ as always |
$=9+\mathrm{12i}\u20134$ | ${\text{(2i)}}^{2}=\text{(2i)}\text{(2i)}=\text{(2)}\text{(2)}\text{(i)}\text{(i)}={\mathrm{4i}}^{2}=\mathrm{\u20134}$ |
$=5+\mathrm{12i}$ | 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 $a\u20131$ is thrown into the picture.
Example: Simplify $\sqrt{-\text{20}}$ | |
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$\sqrt{-\text{20}}$ = $\sqrt{(4)(5)(-1)}$ | 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.
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