# Rationalizing Complex Expressions

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Algebra Two $\longrightarrow$
Complex Numbers $\longrightarrow$

Objectives
• Rationalize expressions that contain a complex number in the denominator
• Learn how to divide by a complex number
Lesson Description

Dividing by a complex number, or equivalently, working with a fraction that has a complex number denominator, requires rationalization for the final answer. This lesson covers the nuances of dividing by a complex number via rationalization and simplification.

Practice Problems

Practice problems and worksheet coming soon!

## There's No i in.... Denominator?

Because the imaginary number $i$ represents $\sqrt{-1}$, the consensus among teachers is that we treat it like a square root when it comes to agreeing on what it means to "simplify" an expression. As we saw with rationalizing square roots » previously, one of the biggest agreed-upon no-nos for teachers is leaving answers with radicals in the denominator, and because of that, we will also not be allowed to leave $i$ in the denominator.There's another good reason for this, however. As we'll see throughout the lesson, any number fraction with a complex number or pure imaginary denominator can be expressed as a single complex number $a + bi$ where $a$ and $b$ are real numbers (but possibly fractions).

## Pure Imaginary

The simplest case for simplifying fractions with imaginary number denominators is to have a purely imaginary number rather than a complex number ». Whether that denominator is just $i$ or a multiple of $i$, we will rationalize it the same way in this case: multiply by $i/i$.

Example 1Rationalize.$$\frac{-6+2i}{i}$$$\blacktriangleright$ Multiply this fraction by $i/i$, which will preserve its value due to the fact that we are multiplying by a form of $1$, and will also create a real number denominator.$$\frac{-6+2i}{i} \cdot \frac{i}{i}$$$$\longrightarrow \frac{-6i + 2i^2}{i^2} \longrightarrow \frac{-2 - 6i}{-1}$$$$\longrightarrow 2+ 6i$$

Pro Tip
All rationalizing happens this way when the denominator of the complex number fraction you are rationalizing is $i$ or a multiple of $i$. Multiply top and bottom by $i/i$ and simplify using the properties of the imaginary number.

## Complex Denominators

Recall that purely imaginary numbers are just a subset of complex numbers ». If you have a complex number denominator, the process for rationalizing is similar to the approach for rationalizing mixed radical denominators » which we looked at previously.
Rationalizing a+bi DenominatorsA complex number is of the form $a + bi$ where $a$ and $b$ are each real numbers. The conjugate of $a+bi$ is defined to be $a-bi$.For any fraction with a complex denominator and a real, imaginary, or complex numerator, rationalization is achieved by multiplying by the conjugate of the denominator.$$\frac{c}{a+bi} \cdot \frac{a-bi}{a-bi}$$or$$\frac{di}{a+bi} \cdot \frac{a-bi}{a-bi}$$or$$\frac{c+di}{a+bi} \cdot \frac{a-bi}{a-bi}$$for real numbers $c$ and $d$.
Depending on the numerator, we may be distributing or applying "FOIL". Let's see an example of each.

Example 2Rationalize the following fraction.$$\frac{i}{-1+4i}$$$\blacktriangleright$ The denominator is a complex number, so we will rationalize this fraction by multiplying by the conjugate of the denominator.$$\frac{i}{-1+4i} \cdot \frac{-1 - 4i}{-1 - 4i}$$$$\frac{i(-1-4i)}{(-1+4i)(-1-4i)}$$$$\frac{-i-4i^2}{1 + 4i - 4i -16i^2}$$$$\frac{4 - i}{17}$$

Example 3Rationalize the following fraction.$$\frac{2-5i}{3+i}$$$\blacktriangleright$ Once again, we need to multiply by the conjugate of the denominator.$$\frac{2-5i}{3+i} \cdot \frac{3 - i}{3 - i}$$$$\frac{6 - 2i -15i + 5i^2}{9 - 3i + 3i - i^2}$$$$\frac{1 - 17i}{10}$$
You Should Know
Always read your instructions! Some questions want you to rationalize these types of fractions and turn in answers in the form of a fraction, while other questions ask you to turn in answers in the form $a + bi$. Even though the work process is the same, the answer forms are slightly different.For instance, the answer to Example 3 above would be$$\frac{1}{10} - \frac{17i}{10}$$if the instructions had instead said "Rationalize the following fraction and express your answer in the form $a + bi$". And in that case, if you turned in the answer$$\frac{1 - 17i}{10}$$and your teacher is a mean, they'll dock you points!

## This is Division

You may have wondered in the prior lesson about complex number arithmetic » why we looked at addition, subtraction, and multiplication but not division. Remember that fractions are really just shorthand notation for division, and so when you're given a problem such as one we've seen in this lesson, you're being asked to divide.For instance, when you're asked to simplify$$\frac{2-5i}{3+i}$$you're also answering the question "what is $(2-5i) \div (3+i)$?"Rationalizing a complex number fraction isn't just your job for the task of "simplifying", but it's literally how you can divide one complex number by another and get a single $a+bi$ answer! From this example, we can see that$$(2-5i) \div (3+i) = \frac{1}{10} - \frac{17i}{10}$$

## Put It To The Test

In each of the following exercises, rationalize or divide. Express your answer as either a fraction with a real number denominator, or a complex number of the form $a + bi$.

Example 4$$\frac{-2}{3i}$$
Show solution
$\blacktriangleright$ Multiply by $i/i$.$$\frac{-2}{3i} \cdot \frac{i}{i}$$$$\frac{-2i}{3i^2}$$$$\frac{2i}{3}$$If you really want to be proper about also reporting the answer in the form $a + bi$, we could say this answer is$$0 + \frac{2}{3} \, i$$Note that Many students are tempted to multiply by $3i/3i$, and while that isn't a wrong or illegal move, it is inefficient.

Example 5$$\frac{60+2i}{7+2i}$$
Show solution
$\blacktriangleright$ We need to multiply the top and bottom of this expression by the conjugate of the denominator.$$\frac{60+2i}{7+2i} \cdot \frac{7-2i}{7-2i}$$$$\frac{420 - 14i + 120i - 4i^2}{49 -14i +14i -4i^2}$$$$\frac{424 + 106i}{53}$$Do be careful to check - this is a tricky one for further simplifying. $53$ goes into each of these numbers!$$\frac{\cancel{53}(8+2i)}{\cancel{53}} = 8 + 2i$$Also, this answer is already in the form $a + bi$.

Example 6$$\frac{9}{6i - 2}$$
Show solution
$\blacktriangleright$ Similar to the last problem, multiply by the conjugate of the denominator.$$\frac{9}{6i - 2} \cdot \frac{6i + 2}{6i + 2}$$$$\frac{54i + 18}{36i^2 + 12i - 12i - 4}$$$$\frac{54i + 18}{-40}$$$$\frac{-9 - 27i}{20}$$Or, in $a + bi$ form,$$-\frac{9}{20} - \frac{27}{20} \, i$$Note that some teachers will try to tell you that you need to rewrite the denominator as $-2 + 6i$, and that the conjugate is really $-2 - 6i$. You will ultimately get an equivalent answer. It happens to be true that the conjugate trick works both ways.

Example 7$$-5 - 10i \div 8 + i$$
Show solution
$\blacktriangleright$ Don't let the division sign fool you - it's equivalent to write this as a fraction, and when we do it will look familiar.$$\frac{-5 - 10i}{8 + i}$$$$\longrightarrow \frac{-5 - 10i}{8 + i} \cdot \frac{8 - i}{8 - i}$$$$\frac{-40 + 5i -80i + 10i^2}{64 - 8i + 8i -i^2}$$$$\frac{-50 - 75i}{65}$$$$\frac{-10 - 15i}{13}$$Or, in $a + bi$ form,$$-\frac{10}{13} - \frac{15}{13} \, i$$

Example 8$$5i + 9 \div 4$$
Show solution
$\blacktriangleright$ We don't actually need to do any rationalizing here - the denominator is not imaginary! Sometimes teachers throw this kind of thing in the mix to trick you.Answer:$$\frac{5i + 9}{4}$$or$$\frac{9}{4} + \frac{5}{4} \, i$$

Lesson Takeaways
• Understand when teachers expect you to rationalize an imaginary expression
• Know how to rationalize purely imaginary denominators efficiently (just $i$, ignore the denominator coefficient)
• Know how to rationalize imaginary expressions with complex number denominators
• Maintain awareness about the way in which your instructions want you to give answers (often specifically asked for $a + bi$ form)
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