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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: 5i
Simplify: i + 5i
Answer: 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 + 5i = i ( 1 + 5 ) = 6i )
Simplify: 2i + 3
Answer: You can't simplify it.

Now let's try something a little more involved.

Example: Simplify the expression (3+2i)2
( 3 + 2i ) 2 = 3 2 + 2 ( 3 ) ( 2i ) + ( 2i ) 2 because ( x + a ) 2 = x 2 + 2 ax + a 2 as always
= 9 + 12i 4 (2i) 2 = (2i) (2i) = (2) (2) (i) (i) = 4i 2 = –4
= 5 + 12i we can combine the 9 and –4, but not the 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 (3+2i) 2 = 5 + 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 20 size 12{ sqrt { - "20"} } {}
20 size 12{ sqrt { - "20"} } {} = ( 4 ) ( 5 ) ( 1 ) size 12{ sqrt { \( 4 \) \( 5 \) \( - 1 \) } } {} as always, factor out the perfect squares
= 4 size 12{ sqrt {4} } {} 5 size 12{ sqrt {5} } {} 1 size 12{ sqrt { - 1} } {} then split it, because ab size 12{ sqrt { ital "ab"} } {} = a size 12{ sqrt {a} } {} b size 12{ sqrt {b} } {}
= 2i 5 size 12{ sqrt {5} } {} 4 size 12{ sqrt {4} } {} =2, 1 size 12{ sqrt { - 1} } {} = i , and 5 size 12{ sqrt {5} } {} is just 5 size 12{ sqrt {5} } {}
Check
Is 2i 5 size 12{ sqrt {5} } {} really the square root of –20? If it is, then when we square it, we should get –20.
( 2i 5 ) 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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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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