Algebra: When Things Get Complex

When Things Get Complex

So far in this section, the vast majority of the radicands you've seen have been positive. They can't help it€”they're just upbeat people, and there's nothing wrong with that. However, you do need to know how to handle it when a bit of negativity creeps in.

Sometimes a negative is no problem. Specifically, if the index of a radical is odd, then a negative radicand is completely valid. For instance,

can be simplified as -2x, since (-2x)(-2x)(-2x) = -8x³. Basically, a negative thing multiplied by itself an odd number of times will also be negative. However, if a radical has an even index, there's trouble.

While the expression ˆš16 is easy to simplify (ˆš16 = 4, since 4 · 4 = 16), the radical expression ˆš-16 is not. What times itself can equal a negative number? If you remember, the only way to multiply two numbers together and get a negative was when the two numbers had different signs, and there's no way something can have a different sign than itself!

There's Something in Your i

Luckily, math people are resilient folks, and they can invent stuff that allows even the most impossible things to happen. One of their handiest inventions is i, a letter representing the solution to a problematic negative radicand. The letter i is short for "imaginary number," and has the value i = ˆš-1. If you think about it, i would have to be imaginary, because no real number could have the value ˆš-1, for all the reasons I laid out above.

A number containing i, such as 2i or -5i, is said to be an imaginary number. Furthermore, any number of the form a + bi (where a and b are real numbers) is said to be complex. Basically, a complex number is made up of an imaginary part, bi, added to or subtracted from a real part, a. For example, in the complex number 4 - 7i, the real part is 4 and the imaginary part is -7i.

Every complex number a + bi has a conjugate paired with it, which is equal to a - bi; in other words, the only difference between a complex number and its conjugate is the sign preceding the imaginary part. For the complex number example I used a few moments ago, 4 - 7i, the conjugate would be 4 + 7i.

Probably the most important thing to remember about complex and imaginary numbers is that i² = -1, since i² = (ˆš-1)²; remember, you just learned that if a radicand is raised to an exponent that matches its index, both disappear, leaving behind only what was beneath the radical (-1 in this case). I know; it's weird that a number squared could be negative, but it's only true for imaginary numbers.

Example 7: Simplify the expressions.

• ˆš-40
• Solution: Rewrite ˆš-40 as

• The perfect square (4 = 2²) can be pulled out of the radical, and ˆš-1 can be rewritten as i: 2iˆš10.
• (b) i⁵
• Solution: Since you know i² = -1, rewrite i⁵ using as many i² factors as possible.
• i⁵ = i² · i² · i
• (The sum of the exponents on the right side, 2 + 2 + 1, must equal 5, the exponent on the left side.) Now replace each i² with -1.
• i⁵ = (-1)(-1)i
• i⁵ = i

Simplifying Complex Expressions

You'll need to know how to add, subtract, multiply, and divide complex numbers, but every complex number is really just a binomial, so you'll apply the same methods in Introducing Polynomials that you used with polynomials (except when it comes to division, that is). Here's a quick rundown describing how the four major operations work with complex numbers:

• Addition: Since imaginary numbers contain the same variable, i, treat them as like terms. Just add up the real parts and the imaginary parts separately.

• (3 - 4i) + (2 + 9i) = 3 + 2 - 4i + 9i = 5 + 5i

• Subtraction: Distribute the negative sign, and all that's left behind is a simple addition problem.

• (3 - 4i) - (2 + 9i) = 3 - 4i -2 - 9i = 1 - 13i

• Multiplication: Just like any product of binomials, distribute each term in the first complex number through the second complex number separately.

• (2 + 3i)(5 - 2i) = 2 · 5 + 2(-2i) + 3i · 5 + 3i(-2i)
• = 10 - 4i + 15i - 6i²
• Replace i² with -1 and combine like terms.
• 10 - 4i + 15i - 6(-1)
• = 10 - 4i + 15i + 6
• = 16 + 11i

• Division: Good news! You don't have to do long or synthetic division to calculate the quotient of complex numbers. To calculate (1 - i) · (2 + 7i), start by writing the quotient as a fraction.

• Multiply both the numerator and denominator by the conjugate of the denominator.

• Multiply the numerators together and write the result over the product of the denominators.

• If you want, you can write each term of the numerator divided separately by the denominator to get the result into the official a + bi form of a complex number.

Remember, no simplified complex number will ever contain an i2 in it; you should always replace that with -1 and simplify like terms.

Excerpted from The Complete Idiot's Guide to Algebra © 2004 by W. Michael Kelley. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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