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# 4.7: Products and Quotients of Complex Numbers

Difficulty Level: At Grade Created by: CK-12
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Practice Products and Quotients of Complex Numbers

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Working with complex numbers can be decidedly strange. It is natural to wonder why we put so much time and effort into learning how to properly manipulate numbers that don't even exist! There are, however, many real and rather fun and interesting applications of complex numbers in the "real world."

This website: http://www.picomonster.com/ has a fascinating interactive application using complex numbers. Check it out.

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### Guidance

In this lesson, we will explore the multiplication and division of complex numbers. In general, operations on complex numbers behave just like operations on regular polynomials: combine like terms, FOIL binomials, distribute, and so on. The differences are generally apparent when the products and quotients of complex numbers are simplified.

Recall the unusual case of i2\begin{align*}i^2\end{align*}:

\begin{align*}i = \sqrt{-1} \therefore i^2 = (\sqrt{-1})^2 = -1\end{align*}

Since \begin{align*}i^2 = -1\end{align*}, multiplying complex numbers often results in imaginary terms becoming real terms.

Multiplying Complex Numbers

When multiplying complex numbers in rectangular form, recall the method for multiplying two binomials (sometimes called FOIL): (m + n)(x + y) = mx + my + nx + ny. We use the same procedure for multiplying complex numbers:

(a + bi) + (c + di) = ac + adi + bci + bdi2

But, unlike the algebraic expression, the above expression contains the number, i.

Recall that i2 = -1, so bdi2 = bd(-1) = -bd and

adi + bci can be combined and then factored as (ad + bc)i. Thus we have the general result,

(a + bi) + (c + di) = (ac - bd) + (ad + bc)i

Dividing Complex Numbers

To divide two complex numbers is similar to dividing two irrational numbers. Recall that in that problem, the procedure was to find the irrational conjugate of the denominator and then multiply both the numerator and the denominator by that conjugate, for example:

Divide: \begin{align*}\frac{3} {1 + \sqrt{2}}\end{align*}

First find the irrational conjugate of the denominator: \begin{align*}1 - \sqrt{2}\end{align*}, then multiply both the numerator and the denominator by this value:

\begin{align*}\frac{3}{1 + \sqrt{2}} \times \frac{1 - \sqrt{2}} {1 - \sqrt{2}} = \frac{3 - 3\sqrt{2}} {1 - 2}\end{align*}

this reduces to

\begin{align*}= \frac{3 - 3\sqrt{2}} {-1}\end{align*}

or

\begin{align*}= -3 + 3\sqrt{2}\end{align*}

The process is very much the same with complex numbers. With complex numbers, since you are interested in eliminating the complex numbers from the denominator, you find the complex conjugate of the denominator and multiply BOTH the numerator AND the denominator by it.

You find the complex conjugate in the same way you found the conjugate of irrational numbers, change the sign of the imaginary part. For instance, the complex conjugate of 4 + 3i is 4 – 3i

A complex number multiplied by its complex conjugate will yield a real number. By recalling (a + b)(a - b) = a2 - b2 simplifying a complex number and its conjugate can be easy, for example:

The conjugate of 4 + 3i is found by retaining the real part (4) and reversing only the sign of the imaginary part (that is, 3i becomes -3i)
(4 + 3i)(4 - 3i) = 16 - 12i + 12i - 9i2. Notice that -12i and 12i cancel. Also recall that i2 = -1

That gives us:

16 + 9 = 25

Therefore: (4 + 3i)(4 – 3i) = 25

The product of this complex number and its conjugate is 25.

When multiplying complex numbers sometimes intuition about the nature of the product can mislead.

For example in (a + b)(a + b), where all of the terms are real numbers, no terms of each of the four products will cancel. Some of the terms may be combined. However in (1 + i)(1 - i), where some terms are real numbers and some terms are imaginary numbers, this is no longer true. Two of these terms cancel: the first product yields 1 while the last product yields i2 or -1, and those terms cancel!

#### Example A

Multiply: \begin{align*}(6 + 3i) (2 - 3i)\end{align*}

Solution

\begin{align*}(6 + 3i)(2 - 3i) = 12 - 18i + 6i - 9i^2\end{align*}

\begin{align*}= 21 - 12i\end{align*}

(Note when combining like terms: -9i2 reduces to -9(-1) or 9)

#### Example B

Multiply: \begin{align*}(5 - 7i) (5 + 7i)\end{align*}

Solution

\begin{align*}(5 - 7i)(5 + 7i) = 25 + 35i - 35i - 49i^2\end{align*}

or

\begin{align*}= 25 + 0 - 49(-1)\end{align*}

\begin{align*}= 74\end{align*}

#### Example C

Find the quotient: \begin{align*}\frac{6 - 3i} {4 + 3i}\end{align*}

Solution

First, observe that the complex conjugate of the denominator is 4 – 3i

Multiply both the numerator and the denominator by 4 – 3i: \begin{align*}\frac{6 - 3i} {4 + 3i} \times \frac{4 - 3i} {4 - 3i} = \frac{24 - 18i + 12i + 9i^2} {16 - 12i + 12i - 9i^2}\end{align*}

\begin{align*}= \frac{15 - 6i} {25}\end{align*}

\begin{align*}\frac{6 - 3i} {4 + 3i} = \frac{15 - 6i} {25}\end{align*}

### Vocabulary

Conjugates are binomial terms which are equal aside from inverse operations between them, e.g. (3 + 2x) and (3 - 2x).

Complex conjugates such as: (3 + 2i) and (3 - 2i) result in real numbers when multiplied.

### Guided Practice

1) Multiply \begin{align*}(6+7i )(2+3i)\end{align*}

2) Multiply \begin{align*}(3 + 6i)(2 - 8i)\end{align*}

3) Divide \begin{align*}\frac{5 + 2i}{7 + 4i}\end{align*}

4) Divide \begin{align*}\frac{4 + i}{5 - i}\end{align*}

Solutions

1) To multiply \begin{align*}(6 + 7i ) \cdot (2 + 3i)\end{align*} treat each complex number as a binomial and apply the FOIL method to find the product.

\begin{align*}6 \cdot 2 = 12\end{align*} multiply the First terms
\begin{align*}6 \cdot 3i = 18i\end{align*} multiply the Outer terms
\begin{align*}7i \cdot 2 = 14i\end{align*} multiply the Inner terms
\begin{align*}7i \cdot 3i = -21\end{align*} multiply the Last terms and simplify the \begin{align*}i^2\end{align*}

\begin{align*}12 +18i + 14i - 21 \to 12 + 32i -21 \to -9 + 32i\end{align*} Combine like terms and simplify

\begin{align*}\therefore (6 + 7i) \cdot (2 + 3i) = -9 + 32i\end{align*}

2) To multiply \begin{align*}(3 + 6i) \cdot (2 - 8i)\end{align*} treat each complex number as a binomial and apply the FOIL method to find the product.

\begin{align*}3 \cdot 2 = 6\end{align*} multiply the First terms
\begin{align*}3 \cdot -8i = -24i\end{align*} multiply the Outer terms
\begin{align*}6i \cdot 2 = 12i\end{align*} multiply the Inner terms
\begin{align*}6i \cdot -8i = +48\end{align*} multiply the Last terms and simplify the \begin{align*}i^2\end{align*}

\begin{align*}6 -24i + 12i + 48 \to 6 - 12i + 48 \to 54 - 12i\end{align*} Combine like terms and simplify

\begin{align*}\therefore (3 + 6i) \cdot (2 - 8i) = 54 - 12i\end{align*}

3) To divide divide \begin{align*}\frac{5 + 2i}{7 + 4i}\end{align*} first determine the conjugate of the denominator \begin{align*}(7 + 4i)\end{align*} the conjugate is the same two terms with the opposite sign between them, in this case \begin{align*}(7 - 4i)\end{align*}

\begin{align*}\left(\frac{5 +2i}{7 + 4i} \right) \bullet \left(\frac{7 - 4i}{7 - 4i} \right)\end{align*} Multiply the numerator and denominator by the conjugate
\begin{align*}\frac{35 + 14i - 20i - 8i^2}{49 - 28i + 28i - 16i^2}\end{align*} FOIL
\begin{align*}\frac{35 - 6i + 8}{49 + 16}\end{align*} Simplify
\begin{align*}\frac{43 - 6i}{65}\end{align*} Simplify

\begin{align*}\therefore \frac{5 + 2i}{7 + 4i} = \frac{43 - 6i}{65}\end{align*}

4) To divide \begin{align*}\frac{4 + i}{5 - i}\end{align*} first identify the conjugate of the denominator: \begin{align*}(5 + i)\end{align*}

\begin{align*}\left(\frac{4 + i}{5 - i} \right) \bullet \left(\frac{5 + i}{5 + i} \right)\end{align*} Multiply the numerator and denominator by the conjugate
\begin{align*}\frac{20 + 4i + 5i + i^2}{25 + 5i - 5i - i^2}\end{align*} FOIL
\begin{align*}\frac{35 + 9i - 1}{25 + 1}\end{align*} Simplify
\begin{align*}\frac{34 + 9i}{26}\end{align*} Simplify

\begin{align*}\therefore \frac{4 + i}{5 - i} = \frac{34 + 9i}{26}\end{align*}

### Practice

1. \begin{align*}(-6i) + (-4i)\end{align*}
2. \begin{align*}(12i + 1) + (6i + 11)\end{align*}

Multiply Complex Numbers

1. \begin{align*}-7i(7i + 5)\end{align*}
2. \begin{align*}(2i + 3)^2\end{align*}
3. \begin{align*}(-2i)(-11i)(-12i)\end{align*}
4. \begin{align*}(8i - 5)(11i^3 - 9)\end{align*}
5. \begin{align*}(9\sqrt{-9}) \cdot (14\sqrt{-4})\end{align*}
6. \begin{align*}(10\sqrt{-1}) \cdot (24 \sqrt{-144})\end{align*}
7. \begin{align*}(18 \sqrt{-49}) \cdot (5 \sqrt{-144})\end{align*}

Divide Complex Numbers

1. \begin{align*}\frac{11i - 2}{i + 1}\end{align*}
2. \begin{align*}\frac{10i + 10}{-8i - 10}\end{align*}
3. \begin{align*}\frac{7i + 12}{-12i - 8}\end{align*}
4. \begin{align*}\frac{10i + 5}{-i - 7}\end{align*}
5. \begin{align*}\frac{4i^2 + 4}{-6i^2 - 1}\end{align*}
6. \begin{align*}\frac{5i^2 - 5}{-8i^2 + 3}\end{align*}
7. \begin{align*}\frac{-4\sqrt{-60}}{-2\sqrt{-20}}\end{align*}
8. \begin{align*}\frac {-100\sqrt{-154}}{10\sqrt{-22}}\end{align*}
9. \begin{align*}\frac {48 \sqrt{-170}}{8 \sqrt{-17}}\end{align*}

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### Vocabulary Language: English

Complex Conjugate

Complex conjugates are pairs of complex binomials. The complex conjugate of $a+bi$ is $a-bi$. When complex conjugates are multiplied, the result is a single real number.

complex number

A complex number is the sum of a real number and an imaginary number, written in the form $a + bi$.

Conjugates

Conjugates are pairs of binomials that are equal aside from inverse operations between them, e.g. $(3 + 2x)$ and $(3 - 2x)$.

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