Find the product. Express your answer in standard form. 2 i ( 9 − 4 i ) ( 6 + 5 i ) 2i\left(9-4i\right)\left(6+5i\right)

− 42 + 148 i -42+148i − 42 + 148 i

Find the quotient. Express your answer in standard form. 6 + i 4 − 2 i \frac{6+i}{4-2i}

11 10 + 4 5 i \frac{11}{10}+\frac45i 10 11 + 5 4 i

6 5 + 4 5 i \frac65+\frac45i 5 6 + 5 4 i

11 10 − 4 5 i \frac{11}{10}-\frac45i 10 11 − 5 4 i

Find the quotient. Express your answer in standard form. − 5 + 3 i − 7 − 4 i \frac{-5+3i}{-7-4i}

23 65 − 41 65 i \frac{23}{65}-\frac{41}{65}i 65 23 − 65 41 i

3 5 + 4 5 i \frac35+\frac45i 5 3 + 5 4 i

11. Roots, Radicals, and Complex Numbers

Multiplying and Dividing Complex Numbers

11. Roots, Radicals, and Complex Numbers

Multiplying and Dividing Complex Numbers: Videos & Practice Problems

Video Lessons

Topic summary

Multiplying complex numbers involves using FOIL or distribution, simplifying $i^{2}$ to $- 1$ , and expressing results in standard form $a + b i$ . The complex conjugate, found by reversing the sign of the imaginary part, helps divide complex numbers by eliminating $i$ from denominators. Powers of $i$ cycle every four exponents: $i, -1, -i, 1$ , enabling simplification of high exponents using division by 4 and remainders. These concepts integrate polynomial operations, exponents, and quotients for comprehensive complex number manipulation.

concept

Multiplying Complex Numbers

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Multiplying Complex Numbers Video Summary

Complex numbers can be multiplied using methods similar to those used for algebraic expressions, specifically through distribution or the FOIL (First, Outside, Inside, Last) method. When multiplying complex numbers, it is essential to remember that the imaginary unit $ i $ satisfies the equation $ i^2 = -1 $, which will be crucial for simplifying the results.

To illustrate the multiplication of complex numbers, consider the expression $ 3i \times (7 - 2i) $. The first step involves distributing $ 3i $ across the terms in the parentheses. This results in:

$ 3i \times 7 = 21i $

$ 3i \times (-2i) = -6i^2 $

Next, substituting $ i^2 $ with $-1$ gives:

$ -6i^2 = -6 \times -1 = 6 $

Combining these results leads to:

$ 21i + 6 $

To express the answer in standard form $ a + bi $, it is rearranged to $ 6 + 21i $.

In another example, multiplying $ (-6 + 2i) $ by $ (3 + 4i) $ using the FOIL method yields:

First: $ -6 \times 3 = -18 $

Outside: $ -6 \times 4i = -24i $

Inside: $ 2i \times 3 = 6i $

Last: $ 2i \times 4i = 8i^2 $

Substituting $ i^2 $ gives:

Thus, $ 8i^2 = 8 \times -1 = -8 $

Combining all terms results in:

$ -18 - 8 - 24i + 6i = -26 - 18i $

Finally, this expression is already in standard form $ a + bi $, confirming the final answer as $ -26 - 18i $.

Through these examples, it is clear that multiplying complex numbers involves careful application of distribution or FOIL, simplification using the property of $ i^2 $, and combining like terms to achieve the final result in standard form.

Problem

Perform the indicated operation. Express your answer in standard form.
$open paren 3 plus 8 i close paren squared$

$-55+48i$

$9+64i$

$24i$

$9+24i$

Problem

Find the product. Express your answer in standard form.
$2 i (9 - 4 i) (6 + 5 i)$

$8+18i$

$54-20i$

$54-40i$

$-42+148i$

Problem

Multiply the following and simplify.
$(5 - i) (12)$

$60 plus 12 i$

$60 minus 12 i$

$5 minus 12 i$

$60 i minus 12$

Problem

Multiply the following and simplify.
$13 i times 17 i$

$221$

$221 i$

$negative 221$

$negative 221 i$

Problem

Multiply the following and simplify.
$(7 + 3 i) (7 - 3 i)$

$58$

$40$

$49 plus 9 i$

$49 minus 9 i squared$

concept

Complex Conjugates

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Complex Conjugates Video Summary

Understanding the concept of the complex conjugate is essential for dividing complex numbers. The complex conjugate of a complex number is formed by reversing the sign of its imaginary part. For a complex number expressed as $ a + bi $, where $ a $ is the real part and $ b $ is the imaginary part, the complex conjugate is $ a - bi $. Conversely, if you start with $ a - bi $, its conjugate will be $ a + bi $.

For example, the complex conjugate of $ 1 + 2i $ is $ 1 - 2i $, and the conjugate of $ 1 - 2i $ is $ 1 + 2i $. Similarly, for $ -1 + 2i $, the complex conjugate is $ -1 - 2i $. This process of finding the conjugate is straightforward: simply change the sign of the imaginary component while keeping the real part unchanged.

When you multiply a complex number by its conjugate, the result is always a real number. For instance, multiplying $ 2 + 3i $ by its conjugate $ 2 - 3i $ involves using the FOIL method:

1. First: $ 2 \times 2 = 4 $

2. Outside: $ 2 \times -3i = -6i $

3. Inside: $ 3i \times 2 = 6i $

4. Last: $ 3i \times -3i = -9i^2 $

Since $ i^2 = -1 $, the term $ -9i^2 $ becomes $ +9 $. Combining all these results gives:

$ 4 - 6i + 6i + 9 = 4 + 9 = 13 $

This shows that $ (2 + 3i)(2 - 3i) = 13 $, a real number. In general, multiplying $ a + bi $ by $ a - bi $ yields $ a^2 + b^2 $. This relationship is not only useful for calculations but also serves as a helpful shortcut when working with complex numbers.

In summary, the complex conjugate is a fundamental concept that simplifies operations involving complex numbers, particularly division, which will be explored further in subsequent lessons.

Problem

Find the product of the given complex number and its conjugate.
$4 minus 5 i$

$16$

$41$

$25$

$20$

Problem

Find the product of the given complex number and its conjugate.
$- 7 - i$

$50$

$14$

$49$

$1$

concept

Dividing Complex Numbers

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Dividing Complex Numbers Video Summary

Dividing complex numbers involves a systematic approach to eliminate the imaginary unit from the denominator. When faced with a fraction where the denominator contains an imaginary component, the goal is to simplify it into a form that is easier to work with. This is achieved by using the complex conjugate.

The complex conjugate of a complex number $ a + bi $ is $ a - bi $. For example, if we want to divide $ \frac{3}{1 + 2i} $, we first identify the complex conjugate of the denominator, which is $ 1 - 2i $. To eliminate the imaginary unit from the denominator, we multiply both the numerator and the denominator by this complex conjugate:

\[\frac{3}{1 + 2i} \cdot \frac{1 - 2i}{1 - 2i} = \frac{3(1 - 2i)}{(1 + 2i)(1 - 2i)}\]

Next, we expand both the numerator and the denominator. The numerator becomes:

\[3(1 - 2i) = 3 - 6i\]

For the denominator, we apply the FOIL method:

\[(1 + 2i)(1 - 2i) = 1 \cdot 1 + 1 \cdot (-2i) + 2i \cdot 1 + 2i \cdot (-2i) = 1 - 2i + 2i - 4i^2\]

Since $ i^2 = -1 $, we can simplify $ -4i^2 $ to $ +4 $. Thus, the denominator simplifies to:

\[1 + 4 = 5\]

Now, we can rewrite the fraction as:

\[\frac{3 - 6i}{5}\]

To separate the real and imaginary parts, we can express this as:

\[\frac{3}{5} - \frac{6}{5}i\]

This final expression, $ \frac{3}{5} - \frac{6}{5}i $, represents the quotient of the original complex division in its simplest form. Thus, the process of dividing complex numbers not only provides a solution but also reinforces the understanding of complex conjugates and their role in simplifying expressions.

Problem

Find the quotient. Express your answer in standard form.
$\frac{6 + i}{4 - 2 i}$

$\frac{11}{10}+\frac45i$

$\frac65+\frac45i$

$\frac{11}{10}-\frac45i$

$22+16i$

Problem

Find the quotient. Express your answer in standard form.
$\frac{- 5 + 3 i}{- 7 - 4 i}$

$\frac35+\frac45i$

$18i$

$23-41i$

$\frac{23}{65}-\frac{41}{65}i$

concept

Powers of i

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Powers of i Video Summary

The imaginary unit $ i $ is defined as the square root of negative one, $ i = \sqrt{-1} $. When raising $ i $ to various powers, a pattern emerges that simplifies calculations significantly. Understanding the first few powers of $ i $ allows us to easily determine higher powers.

Starting with the basic powers:

$ i^1 = i $
$ i^2 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = -1 $
$ i^3 = i^2 \cdot i = -1 \cdot i = -i $
$ i^4 = i^2 \cdot i^2 = (-1) \cdot (-1) = 1 $

Continuing this pattern, we can express higher powers of $ i $ in terms of these four results. For example:

$ i^5 = i^4 \cdot i = 1 \cdot i = i $
$ i^6 = i^4 \cdot i^2 = 1 \cdot (-1) = -1 $
$ i^7 = i^4 \cdot i^3 = 1 \cdot (-i) = -i $
$ i^8 = i^4 \cdot i^4 = 1 \cdot 1 = 1 $

This reveals a repeating cycle every four powers: $ i, -1, -i, 1 $. Therefore, any power of $ i $ can be simplified by finding the remainder when the exponent is divided by 4:

For any integer $ n $:

If $ n \mod 4 = 0 $, then $ i^n = 1 $
If $ n \mod 4 = 1 $, then $ i^n = i $
If $ n \mod 4 = 2 $, then $ i^n = -1 $
If $ n \mod 4 = 3 $, then $ i^n = -i $

This cyclical nature allows for quick calculations of any power of $ i $, making it a powerful tool in complex number arithmetic.

Problem

Evaluate the following powers of $i$ .
$open paren 3 i close paren to the fourth power$

$81$

$negative 81$

$9$

$negative 9$

Problem

Evaluate the following powers of $i$ .
${(4 i)}^{- 3}$

$64 i cubed$

$64 i$

$1 over 64 i$

$1 over 64 i squared$

concept

Higher Powers of i

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Higher Powers of i Video Summary

Understanding the powers of the imaginary unit $ i $ is essential in complex number calculations. The powers of $ i $ cycle through four distinct values: $ i $, $-1$, $-i$, and $1$. This cyclical nature allows for efficient computation of higher powers without the need to manually count through each power.

To simplify calculations, we can express any power of $ i $ in terms of $ i^4 $, which equals $ 1 $. For instance, to evaluate $ i^{20} $, we can rewrite it as $ (i^4)^5 $. Since $ i^4 = 1 $, it follows that $ i^{20} = 1^5 = 1 $.

For powers that are not multiples of four, such as $ i^{22} $, we can still utilize the cyclical pattern. We start by determining how many complete sets of $ i^4 $ fit into the exponent. In this case, $ i^{22} $ can be expressed as $ (i^4)^5 \cdot i^2 $. Since $ i^2 = -1 $, we find that $ i^{22} = 1 \cdot (-1) = -1 $.

To further streamline the process, we can determine the remainder when the exponent is divided by $ 4 $. If the exponent is evenly divisible by $ 4 $, the result is $ 1 $. For example, $ i^{100} $ can be evaluated by checking if $ 100 \div 4 $ has a remainder. Since $ 100 $ is divisible by $ 4 $, $ i^{100} = 1 $.

For exponents that yield a remainder, we can find the equivalent power of $ i $ based on the remainder. For instance, when calculating $ i^{67} $, dividing $ 67 $ by $ 4 $ gives a quotient of $ 16 $ and a remainder of $ 3 $. Thus, $ i^{67} = i^3 $. Knowing that $ i^3 = -i $, we conclude that $ i^{67} = -i $.

In summary, to evaluate powers of $ i $, determine if the exponent is divisible by $ 4 $. If it is, the result is $ 1 $. If not, calculate the remainder and use it to find the corresponding power of $ i $ from the cycle: $ i^1 = i $, $ i^2 = -1 $, $ i^3 = -i $, and $ i^0 = 1 $. This method provides a quick and efficient way to handle complex exponentiation.

Problem

Simplify the power of $i$ .
$i to the 1003rd power$

$i$

$-1$

$-i$

Problem

Simplify the power of $i$ .
$i to the 85th power$

$i$

-1

$-i$

example

Higher Powers of i Example 1

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Higher Powers of i Example 1 Video Summary

When working with powers of the imaginary unit i, it is essential to remember that the powers cycle every four exponents because i⁴ = 1. This means any power of i can be simplified by dividing the exponent by 4 and using the remainder to determine the equivalent lower power. For example, i¹² equals 1 since 12 is divisible by 4, while i¹⁵ simplifies to i³ because 15 divided by 4 leaves a remainder of 3.

To simplify i³, recall that i² = -1, so i³ = i² × i = -1 × i = -i. Therefore, the expression i¹² + i¹⁵ simplifies to 1 - i.

When multiplying powers of i, add the exponents according to the exponent rule: i^a × i^b = i^a+b. For instance, i⁸ × i⁵ = i¹³. Since 13 divided by 4 leaves a remainder of 1, this simplifies to i¹ = i.

For more complex expressions involving addition, subtraction, multiplication, and division of powers of i, simplify each term by reducing the exponents modulo 4. For example, consider the expression:

\[\frac{i^{10} + i^{15}}{i^{9} - i^{12}}\]

First, simplify each power:

i¹⁰: 10 mod 4 = 2, so i¹⁰ = i² = -1
i¹⁵: 15 mod 4 = 3, so i¹⁵ = i³ = -i
i⁹: 9 mod 4 = 1, so i⁹ = i
i¹²: 12 mod 4 = 0, so i¹² = 1

Substituting these back into the expression gives:

\[\frac{-1 - i}{i - 1}\]

To eliminate the imaginary unit from the denominator, multiply numerator and denominator by the complex conjugate of the denominator, which is i + 1:

\[\frac{-1 - i}{i - 1} \times \frac{i + 1}{i + 1} = \frac{(-1 - i)(i + 1)}{(i - 1)(i + 1)}\]

Expanding the denominator using the difference of squares formula:

\[ (i - 1)(i + 1) = i^2 - 1 = -1 - 1 = -2 \]

Expanding the numerator using the distributive property (FOIL):

\[ (-1)(i) + (-1)(1) + (-i)(i) + (-i)(1) = -i - 1 - i^2 - i \]

Recall that i² = -1, so substitute:

\[ -i - 1 - (-1) - i = -i - 1 + 1 - i = -i - i = -2i \]

Thus, the expression simplifies to:

\[ \frac{-2i}{-2} = i \]

This demonstrates how powers of i can be simplified by reducing exponents modulo 4, applying exponent rules, and rationalizing denominators using complex conjugates to eliminate imaginary units from denominators. Mastery of these techniques is essential for simplifying expressions involving imaginary numbers and complex arithmetic.

Here’s what students ask on this topic:

To multiply two complex numbers, use the FOIL method or distribution, just like with algebraic expressions. For example, if you multiply $a + b i$ and $c + d i$ , you multiply each term: first, outside, inside, and last. Remember that $i i^{2} = - 1$ . After multiplying, simplify all $i^{2}$ terms to $- 1$ . Then combine like terms to write the answer in standard form $a + b i$ , where $a$ and $b$ are real numbers. This process ensures your final answer is clear and properly formatted.

The complex conjugate of a complex number $a + b i$ is $a - b i$ . It is found by changing the sign of the imaginary part. The conjugate is important in division because it helps eliminate the imaginary unit $i$ from the denominator. When you multiply a complex number by its conjugate, the result is a real number: $a^{2} + b^{2}$ . This allows you to rewrite the division problem without $i$ in the denominator, making it easier to simplify and express the quotient in standard form.

To divide complex numbers, multiply both the numerator and denominator by the complex conjugate of the denominator. For example, to divide $\frac{3}{1}$ , multiply numerator and denominator by $1 - 2 i$ . Use FOIL to expand the denominator, which becomes a real number because $i^{2} = - 1$ . Then distribute the numerator. Finally, separate the real and imaginary parts by dividing each term in the numerator by the real denominator. Simplify fractions if possible to express the result in standard form $a + b i$ .

The powers of $i$ cycle every four exponents: $i$ , $- 1$ , $- i$ , and $1$ . For example, $i^{1} = i$ , $i^{2} = - 1$ , $i^{3} = - i$ , and $i^{4} = 1$ . To simplify high powers like $i^{100}$ , divide the exponent by 4 and use the remainder to find the equivalent power. Since 100 is divisible by 4, $i^{100} = 1$ . This method avoids lengthy calculations and quickly simplifies powers of $i$ .

When dividing complex numbers, the denominator often contains an imaginary part, which is difficult to simplify. Multiplying both numerator and denominator by the complex conjugate of the denominator removes the imaginary unit $i$ from the denominator. This works because multiplying a complex number by its conjugate results in a real number, specifically $a^{2} + b^{2}$ . This process rationalizes the denominator, making the expression easier to simplify and express in standard form $a + b i$ . It is a crucial step in dividing complex numbers effectively.

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Multiplying and Dividing Complex Numbers: Videos & Practice Problems

Multiplying Complex Numbers

Multiplying Complex Numbers Video Summary

Perform the indicated operation. Express your answer in standard form.
$open paren 3 plus 8 i close paren squared$

Find the product. Express your answer in standard form.
$2 i (9 - 4 i) (6 + 5 i)$

Multiply the following and simplify.
$(5 - i) (12)$

Multiply the following and simplify.
$13 i times 17 i$

Multiply the following and simplify.
$(7 + 3 i) (7 - 3 i)$