# Connexions

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Textbook by: Kenny M. Felder. E-mail the author

# Playing with i

Module by: Kenny M. Felder. E-mail the author

Summary: 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.

Table 1
A few very simple examples of expressions involving ii
Simplify: i5i5
Simplify: i+5ii+5i
Answer: 6i6i (Add anything to 5 of itself, and you get 6 of it. Or, you can think of this as “pulling out” an ii as follows: i+5i=i(1+5)=6ii+5i=i(1+5)=6i)
Simplify: 2i+32i+3

Now let's try something a little more involved.

Table 2
Example: Simplify the expression (3+2i)2
(3+2i)2 = 32+2(3)(2i)+(2i)2(3+2i)2=32+2(3)(2i)+(2i)2 because(x+a)2=x2+2ax+a2(x+a)2=x2+2ax+a2as always
= 9+12i4=9+12i4 (2i)2=(2i)(2i)=(2)(2)(i)(i)=4i2 =–4(2i)2=(2i)(2i)=(2)(2)(i)(i)=4i2=–4
= 5+12i=5+12i we can combine the 9 and –4, but not the 12i12i.

It is vital to remember that ii is not a variable, and this is not an algebraic generalization. You cannot plug i=3i=3 into that equation and expect anything valid to come out. The equation (3+2i)2=5+12i(3+2i)2=5+12i has been shown to be true for only one number: that number is ii, 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 a1 is thrown into the picture.

Table 3
Example: Simplify 2020 size 12{ sqrt { - "20"} } {}
2020 size 12{ sqrt { - "20"} } {} = (4)(5)(1)(4)(5)(1) size 12{ sqrt { $$4$$ $$5$$ $$- 1$$ } } {} as always, factor out the perfect squares
= 44 size 12{ sqrt {4} } {}55 size 12{ sqrt {5} } {}11 size 12{ sqrt { - 1} } {} then split it, becauseabab size 12{ sqrt { ital "ab"} } {}= aa size 12{ sqrt {a} } {}bb size 12{ sqrt {b} } {}
=2i=2i 55 size 12{ sqrt {5} } {} 44 size 12{ sqrt {4} } {}=2, 11 size 12{ sqrt { - 1} } {}=i=i, and55 size 12{ sqrt {5} } {}is just55 size 12{ sqrt {5} } {}
Check
Is 2i2i 55 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 = 22i2 52 = 4 *-1 * 5 = -20(2i 5)2=22i252=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 ii as the square root of –1. But we see that, using ii, we can now take the square root of any negative number.

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