# 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 itthey'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 -2*x*, since (-2*x*)(-2*x*)(-2*x*) = -8*x*^{3}. 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 16 is easy to simplify (16 = 4, since 4 · 4 = 16), the radical expression -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* = -1. If you think about it, *i* would have to be imaginary, because no real number could have the value -1, for all the reasons I laid out above.

A number containing *i*, such as 2*i* or -5*i*, 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 - 7*i*, the real part is 4 and the imaginary part is -7*i*.

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 - 7*i*, the conjugate would be 4 + 7*i*.

Probably the most important thing to remember about complex and imaginary numbers is that i^{2} = -1, since *i*^{2} = (-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.

##### Talk the Talk

An **imaginary number**, *bi*, is the product of a real number *b* and the imaginary piece *i* = -1. A **complex number** has the form *a* + *bi*, where *a* and *b* are both real numbers; every complex number is paired with a **conjugate**, *a* - *bi*, which matches the complex number exactly, except for the sign preceding *bi*.

##### How'd You Do That?

Every imaginary number is automatically a complex number as well. For instance, 3*i* has form *a* + *bi* (making it complex) if *a* = 0 and *b* = 3. Additionally, every real number is automatically complex as well. Consider the real number 12; it has complex form *a* + *bi* if *a* = 12 and *b* = 0.

- -40
**Solution**: Rewrite -40 as

- The perfect square (4 = 2
^{2}) can be pulled out of the radical, and -1 can be rewritten as*i*: 2*i*10. - (b)
*i*^{5} **Solution**: Since you know*i*^{2}= -1, rewrite*i*^{5}using as many*i*^{2}factors as possible.*i*^{5}=*i*^{2}·*i*^{2}·*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
*i*^{2}with -1. *i*^{5}= (-1)(-1)*i**i*^{5}=*i*

##### You've Got Problems

Problem 7: Simplify the expression -36 + *i*^{3}.

### 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 - 4
*i*) + (2 + 9*i*) = 3 + 2 - 4*i*+ 9*i*= 5 + 5*i*

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

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

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

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

**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 + 7*i*), start by writing the quotient as a fraction.

- 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.

##### You've Got Problems

Problem 8: Given the complex numbers *c* = 3 - 4*i* and *d* = 8 + *i*, calculate (a) *c* + *d*, (b) *c* - *d*, (c) *c* · *d*, and (d) *c* · *d*.

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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**See also:**