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Operations on Complex Numbers (page 2 of 3)

Sections: Introduction, Operations with complexes, The Quadratic Formula

Complex numbers are "binomials" of a sort, and are added, subtracted, and multiplied in a similar way. (Division, which is further down the page, is a bit different.)

• Simplify (2 + 3i) + (1 – 6i).

(2 + 3i) + (1 – 6i) = (2 + 1) + (3i – 6i) = 3 + (–3i) = 3 – 3i

• Simplify (5 – 2i) – (–4 – i).

(5 – 2i) – (–4 – i)

= (5 – (–4)) + (–2i – (–i)) = (5 + 4) + (–2i + i)

= (9) + (–1i) = 9 – i

• Simplify (2 – i)(3 + 4i).

(2 – i)(3 + 4i) = (2)(3) + (2)(4i) + (–i)(3) + (–i)(4i)

= 6 + 8i – 3i – 4i² = 6 + 5i – 4(–1)

= 6 + 5i + 4 = 10 + 5i

For this last example, if you learned "FOIL", FOILing works for this kind of multiplication. But whatever method you use, remember that multiplying and adding with complexes works just like multiplying and adding polynomials, except that, while x² is just x², i² is –1. That is, you can use the exact same techniques for simplifying complex-number expressions, but you can simplify even further with complexes than with polynomials, because i² reduces to the number –1.

Adding and multiplying complexes isn't too bad. It's when you start on fractions (that is, division) that things turn ugly. Most of the reason for this ugliness is actually arbitrary. Remember back in elementary school, when you first learned fractions? Your teacher would get her panties in a wad if you used "improper" fractions. For instance, you couldn't say " ³/₂ "; you had to convert it to "1 ¹/₂". But now that you're in algebra, nobody cares, and you've probably noticed that "improper" fractions are often more useful than "mixed" fractions. The problem in the case of complexes is that your professor will get his boxers in a bunch if you leave imaginaries in the denominator. So how do you handle this?

Suppose you have the following problem:

• Simplify 

The point here is that they want you to get rid of the i underneath. The 2 is fine, but the i has got to go. To do this, you use the fact that i² = –1. If you multiply top and bottom by i, then the i underneath will vanish in a puff of negativity:

So the answer is –( ³/₂ )i.

This was simple enough, but what if you have something more complicated?

• Simplify

If you multiply top and bottom by i, you get:

Since you still have an i underneath, this didn't help much. So how do you handle this simplification? You use something called "conjugates". The conjugate of a complex number a + bi is the same number, but with the opposite sign in the middle: a – bi. When you multiply conjugates, you are, in effect, multiplying to a difference of squares:

Note that the i's disappeared. This is what the conjugate, difference-of-squares thing is for. Here's how it is used:   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

So the answer is ⁶/₅ – ( ³/₅)i.

In the last step, note how the fraction was split into two pieces. This is because, technically speaking, a complex number is in two parts, the real part and the i part. They aren't supposed to "share" the denominator.

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