Quadratic Functions and Their Algebra Review Questions Answer Key

Learning Objectives

In this section, you will:

• Add together and subtract complex numbers.
• Multiply and separate complex numbers.
• Simplify powers of $i$

Figure ane

Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Ready is one of the nearly recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple.

In social club to better understand it, we demand to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set up of positive integers. The set of rational numbers, in turn, fills a void left by the set up of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this department, we will explore a set of numbers that fills voids in the ready of real numbers and find out how to work inside it.

Expressing Foursquare Roots of Negative Numbers as Multiples of $i$

We know how to find the foursquare root of whatsoever positive real number. In a similar way, we can find the square root of any negative number. The deviation is that the root is not existent. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number $i$ is defined as the square root of $\mathrm{-one.}$

$$\sqrt{\mathrm{-1}}=i$$

Then, using properties of radicals,

$$i^2={\left(\sqrt{-\mathrm{one}}\right)}^{\mathrm{two}}=\mathrm{-1}$$

We tin write the foursquare root of any negative number equally a multiple of $i.$ Consider the square root of $\mathrm{-49.}$

$$\begin{array}{ccc}\hfill \sqrt{\mathrm{-49}}& =& \hfill \sqrt{49\cdot (\mathrm{-1})}\\ & =& \sqrt{49}\sqrt{\mathrm{-1}}\hfill \\ & =& 7i\hfill \end{array}$$

We utilise $7i$ and not $\mathrm{-seven}i$ because the principal root of $49$ is the positive root.

A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written $a+bi$ where $a$ is the real role and $b$ is the imaginary office. For example, $\mathrm{v}+2i$ is a complex number. And so, too, is $3+4i\sqrt{3}.$

Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the issue is also a positive real number. Complex numbers consist of real and imaginary numbers.

Imaginary and Complex Numbers

A circuitous number is a number of the form $a+bi$ where

• $a$ is the real role of the circuitous number.
• $b$ is the imaginary part of the complex number.

If $b=0,$ and so $a+bi$ is a real number. If $a=0$ and $b$ is not equal to 0, the circuitous number is called a pure imaginary number. An imaginary number is an even root of a negative number.

How To

Given an imaginary number, express it in the standard grade of a complex number.

1. Write $\sqrt{-a}$ equally $\sqrt{a}\sqrt{\mathrm{-1}}.$
2. Express $\sqrt{\mathrm{-1}}$ as $i.$
3. Write $\sqrt{a}\cdot i$ in simplest course.

Example 1

Expressing an Imaginary Number in Standard Class

Express $\sqrt{\mathrm{-9}}$ in standard form.

Attempt Information technology #i

Limited $\sqrt{\mathrm{-24}}$ in standard form.

Plotting a Complex Number on the Complex Plane

Nosotros cannot plot complex numbers on a number line as nosotros might existent numbers. However, we can still represent them graphically. To represent a complex number, we need to accost the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical centrality represents the imaginary component. Complex numbers are the points on the airplane, expressed as ordered pairs $(a,b),$ where $a$ represents the coordinate for the horizontal axis and $b$ represents the coordinate for the vertical axis.

Let's consider the number $\mathrm{-ii}+\mathrm{iii}i.$ The real part of the complex number is $\mathrm{-2}$ and the imaginary part is three. Nosotros plot the ordered pair $(\mathrm{-two},3)$ to represent the circuitous number $\mathrm{-2}+3i,$ as shown in Figure 2.

Figure 2

Complex Airplane

In the complex airplane, the horizontal axis is the real axis, and the vertical centrality is the imaginary axis, as shown in Figure 3.

Figure 3

How To

Given a complex number, represent its components on the complex airplane.

1. Determine the real function and the imaginary function of the complex number.
2. Move along the horizontal axis to prove the real role of the number.
3. Move parallel to the vertical axis to evidence the imaginary function of the number.
4. Plot the bespeak.

Example two

Plotting a Complex Number on the Complex Airplane

Plot the complex number $3-4i$ on the complex plane.

Try It #2

Plot the complex number $\mathrm{-4}-i$ on the complex aeroplane.

Adding and Subtracting Circuitous Numbers

Merely as with real numbers, we can perform arithmetics operations on complex numbers. To add or subtract complex numbers, we combine the real parts and so combine the imaginary parts.

Complex Numbers: Addition and Subtraction

Adding complex numbers:

$$\left(a+bi\right)+\left(c+di\right)=\left(a+c\right)+\left(b+d\right)i$$

Subtracting complex numbers:

$$\left(a+bi\right)-\left(c+di\right)=\left(a-c\right)+\left(b-d\right)i$$

How To

Given 2 circuitous numbers, find the sum or difference.

1. Place the existent and imaginary parts of each number.
2. Add or decrease the real parts.
3. Add or decrease the imaginary parts.

Instance 3

Calculation and Subtracting Complex Numbers

Add or decrease every bit indicated.

1. ⓐ $\left(3-4i\right)+\left(2+\mathrm{five}i\right)$
2. ⓑ $\left(\mathrm{-5}+7i\right)-\left(\mathrm{-11}+\mathrm{two}i\right)$

Try It #three

Subtract $\mathrm{two}+\mathrm{v}i$ from $\mathrm{three}–4i.$

Multiplying Complex Numbers

Multiplying complex numbers is much similar multiplying binomials. The major difference is that nosotros work with the real and imaginary parts separately.

Multiplying a Complex Number by a Real Number

Lets begin past multiplying a complex number past a real number. We distribute the existent number just as we would with a binomial. Consider, for example, $3\left(6+\mathrm{ii}i\right)$ :

How To

Given a complex number and a existent number, multiply to find the production.

1. Use the distributive property.
2. Simplify.

Example iv

Multiplying a Complex Number by a Real Number

Find the production $4(2+5i).$

Try It #4

Find the product: $\frac{1}{\mathrm{two}}\left(5-2i\right).$

Multiplying Complex Numbers Together

Now, permit'south multiply two complex numbers. We tin utilise either the distributive property or more specifically the FOIL method considering we are dealing with binomials. Recall that FOIL is an acronym for multiplying Starting time, Inner, Outer, and Last terms together. The departure with complex numbers is that when we get a squared term, $i^2,$ it equals $\mathrm{-ane.}$

$$\begin{array}{cccc}\hfill (a+bi)(c+di)& =& ac+adi+bci+bd{i}^2\hfill & \\ & =& ac+adi+bci-bd\hfill & i^2=\mathrm{-1}\hfill \\ & =& (ac-bd)+(ad+bc)i\hfill & \text{Group real terms and imaginary terms}.\hfill \end{array}$$

How To

Given ii complex numbers, multiply to find the product.

1. Use the distributive property or the FOIL method.
2. Remember that $i^2=\mathrm{-i.}$
3. Group together the real terms and the imaginary terms

Instance 5

Multiplying a Circuitous Number by a Circuitous Number

Multiply: $\left(\mathrm{four}+3i\right)(2-\mathrm{five}i).$

Attempt It #5

Multiply: $\left(3-4i\right)\left(\mathrm{two}+\mathrm{three}i\right).$

Dividing Complex Numbers

Dividing two complex numbers is more complicated than adding, subtracting, or multiplying considering we cannot divide by an imaginary number, meaning that whatsoever fraction must accept a real-number denominator to write the answer in standard form $a+bi.$ We demand to detect a term by which we tin multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a existent number every bit the denominator. This term is called the complex cohabit of the denominator, which is institute by changing the sign of the imaginary role of the complex number. In other words, the complex conjugate of $a+bi$ is $a-bi.$ For example, the production of $a+bi$ and $a-bi$ is

$$\begin{array}{ccc}\hfill (a+bi)(a-bi)& =& a^2-abi+abi-b^2{i}^{\mathrm{ii}}\\ & =& a^2+b^2\hfill \end{array}$$

The result is a real number.

Note that circuitous conjugates accept an contrary relationship: The circuitous conjugate of $a+bi$ is $a-bi,$ and the complex conjugate of $a-bi$ is $a+bi.$ Further, when a quadratic equation with existent coefficients has complex solutions, the solutions are always complex conjugates of one another.

Suppose nosotros want to divide $c+di$ past $a+bi,$ where neither $a$ nor $b$ equals goose egg. We first write the division every bit a fraction, then find the complex conjugate of the denominator, and multiply.

$$\frac{c+di}{a+bi}\text{where}a\ne 0\text{and}b\ne 0$$

Multiply the numerator and denominator past the complex conjugate of the denominator.

$$\frac{\left(c+di\right)}{\left(a+bi\right)}\cdot \frac{\left(a-bi\right)}{\left(a-bi\right)}=\frac{\left(c+di\right)\left(a-bi\right)}{\left(a+bi\right)\left(a-bi\right)}$$

Apply the distributive property.

$$=\frac{ca-cbi+adi-bd{i}^{\mathrm{two}}}{a^2-abi+abi-b^{\mathrm{two}}{i}^{\mathrm{ii}}}$$

Simplify, remembering that $i^2=\mathrm{-1.}$

$$\begin{array}{l}=\frac{ca-cbi+adi-bd(\mathrm{-i})}{a^{\mathrm{ii}}-abi+abi-b^2(\mathrm{-ane})}\hfill \\ =\frac{(ca+bd)+(ad-cb)i}{a^2+b^{\mathrm{ii}}}\hfill \end{array}$$

The Complex Conjugate

The circuitous conjugate of a complex number $a+bi$ is $a-bi.$ It is institute by irresolute the sign of the imaginary part of the complex number. The real part of the number is left unchanged.

• When a complex number is multiplied by its complex conjugate, the result is a real number.
• When a complex number is added to its complex conjugate, the issue is a real number.

Example 6

Finding Complex Conjugates

Detect the complex conjugate of each number.

1. ⓐ $\mathrm{two}+i\sqrt{5}$
2. ⓑ $-\frac{1}{2}i$

Analysis

Although we have seen that we can discover the circuitous cohabit of an imaginary number, in practice nosotros more often than not observe the complex conjugates of but complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can merely multiply by $i.$

Try Information technology #6

Find the complex conjugate of $\mathrm{-iii}+\mathrm{iv}i.$

How To

Given two complex numbers, divide one by the other.

1. Write the division problem as a fraction.
2. Make up one's mind the complex conjugate of the denominator.
3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator.
4. Simplify.

Instance 7

Dividing Complex Numbers

Split up: $\left(\mathrm{ii}+\mathrm{five}i\right)$ by $\left(4-i\right).$

Simplifying Powers of i

The powers of $i$ are cyclic. Let'south look at what happens when we raise $i$ to increasing powers.

$$\begin{array}{l}{i}^1=i\hfill \\ i^2=\mathrm{-one}\hfill \\ i^{\mathrm{iii}}=i^2\cdot i=\mathrm{-1}\cdot i=-i\hfill \\ i^{\mathrm{iv}}=i^3\cdot i=-i\cdot i=-i^{\mathrm{two}}=-(\mathrm{-1})=1\hfill \\ i^5=i^4\cdot i=\mathrm{ane}\cdot i=i\hfill \end{array}$$

We can see that when we get to the fifth ability of $i,$ it is equal to the first power. Every bit nosotros continue to multiply $i$ past increasing powers, nosotros will run into a cycle of 4. Let's examine the next four powers of $i.$

$$\begin{array}{l}{i}^6=i^5\cdot i=i\cdot i=i^{\mathrm{two}}=\mathrm{-one}\hfill \\ i^7=i^6\cdot i=i^2\cdot i=i^3=-i\hfill \\ i^8=i^7\cdot i=i^3\cdot i=i^4=1\hfill \\ i^9=i^{\mathrm{viii}}\cdot i=i^4\cdot i=i^5=i\hfill \end{array}$$

The cycle is repeated continuously: $i,\mathrm{-i},-i,1,$ every iv powers.

Case 8

Simplifying Powers of $i$

Evaluate: $i^{35}.$

Try It #vii

Q&A

Can we write $i^{35}$ in other helpful ways?

Every bit we saw in Case 8, we reduced $i^{35}$ to $i^{\mathrm{three}}$ by dividing the exponent by 4 and using the residue to discover the simplified form. But perhaps some other factorization of $i^{35}$ may be more useful. Tabular array 1 shows some other possible factorizations.

Factorization of $i^{35}$	$i^{34}\cdot i$	$i^{33}\cdot i^2$	$i^{31}\cdot i^4$	$i^{19}\cdot i^{16}$
Reduced form	${\left(i^2\right)}^{17}\cdot i$	$i^{33}\cdot \left(\mathrm{-one}\right)$	$i^{31}\cdot 1$	$$i^{\mathrm{xix}}\cdot {\left(i^4\right)}^4$$
Simplified grade	${\left(\mathrm{-1}\right)}^{17}\cdot i$	$-i^{33}$	$i^{31}$	$i^{\mathrm{xix}}$

Tabular array 1

Each of these will eventually result in the answer we obtained to a higher place but may require several more steps than our before method.

2.4 Section Exercises

Verbal

one .

Explicate how to add together complex numbers.

2 .

What is the basic principle in multiplication of complex numbers?

iii .

Give an example to testify that the product of ii imaginary numbers is non e'er imaginary.

4 .

What is a characteristic of the plot of a real number in the complex airplane?

Algebraic

For the following exercises, evaluate the algebraic expressions.

5 .

If $y={\mathrm{ten}}^{\mathrm{ii}}+x-4,$ evaluate $y$ given $x=2i.$

vi .

If $y=x^{\mathrm{three}}-2,$ evaluate $y$ given $x=i.$

vii .

If $y=x^2+3x+\mathrm{v},$ evaluate $y$ given $\mathrm{ten}=2+i.$

8 .

If $y=2x^{\mathrm{two}}+\mathrm{ten}-3,$ evaluate $y$ given $x=2-3i.$

9 .

If $y=\frac{x+1}{\mathrm{ii}-x},$ evaluate $y$ given $x=5i.$

10 .

If $y=\frac{\mathrm{one}+2x}{x+\mathrm{three}},$ evaluate $y$ given $\mathrm{ten}=\mathrm{four}i.$

Graphical

For the following exercises, plot the complex numbers on the complex plane.

Numeric

For the post-obit exercises, perform the indicated performance and express the result as a simplified circuitous number.

15 .

$\left(3+2i\right)+(5-3i)$

16 .

$\left(\mathrm{-two}-4i\right)+\left(\mathrm{one}+6i\right)$

17 .

$\left(\mathrm{-5}+\mathrm{iii}i\right)-(\mathrm{half\; dozen}-i)$

eighteen .

$\left(2-3i\right)-(3+2i)$

xix .

$(\mathrm{-4}+4i)-(\mathrm{-6}+\mathrm{ix}i)$

20 .

$\left(2+\mathrm{three}i\right)(4i)$

21 .

$\left(5-2i\right)(\mathrm{three}i)$

22 .

$\left(6-\mathrm{two}i\right)(5)$

23 .

$\left(\mathrm{-2}+4i\right)\left(\mathrm{eight}\right)$

24 .

$\left(2+3i\right)(4-i)$

25 .

$\left(\mathrm{-1}+2i\right)(\mathrm{-two}+3i)$

26 .

$\left(4-2i\right)(\mathrm{four}+\mathrm{ii}i)$

27 .

$\left(\mathrm{iii}+\mathrm{four}i\right)\left(3-4i\right)$

36 .

$-\sqrt{\mathrm{-iv}}-4\sqrt{\mathrm{-25}}$

Technology

For the following exercises, use a calculator to assist answer the questions.

42 .

Evaluate ${(1+i)}^{\mathrm{yard}}$ for $k=\mathrm{iv},\mathrm{eight},$ and $12$ . Predict the value if $k=16.$

43 .

Evaluate ${(1-i)}^{\mathrm{one\; thousand}}$ for $\mathrm{1000}=2,\mathrm{half\; dozen},$ and $10$ . Predict the value if $k=14.$

44 .

Evaluate ${(\text{l}+i)}^m-{(\text{l}-i)}^{\mathrm{one\; thousand}}$ for $k=4,8,$ and $12$ . Predict the value for $k=\mathrm{sixteen.}$

45 .

Show that a solution of $x^{\mathrm{vi}}+1=0$ is $\frac{\sqrt{\mathrm{iii}}}{2}+\frac{1}{2}i.$

46 .

Testify that a solution of $x^8\mathrm{-1}=0$ is $\frac{\sqrt{\mathrm{ii}}}{\mathrm{two}}+\frac{\sqrt{2}}{2}i.$

Extensions

For the following exercises, evaluate the expressions, writing the consequence as a simplified complex number.

47 .

$\frac{\mathrm{i}}{i}+\frac{4}{i^{\mathrm{iii}}}$

48 .

$\frac{\mathrm{i}}{i^{\mathrm{eleven}}}-\frac{\mathrm{one}}{i^{21}}$

49 .

$i^{\mathrm{seven}}\left(\mathrm{ane}+i^2\right)$

l .

$i^{\mathrm{-3}}+5i^7$

51 .

$\frac{\left(2+i\right)\left(\mathrm{iv}-2i\right)}{(\mathrm{ane}+i)}$

52 .

$\frac{\left(1+\mathrm{iii}i\right)\left(2-4i\right)}{(1+2i)}$

53 .

$\frac{{\left(3+i\right)}^2}{{\left(1+2i\right)}^{\mathrm{two}}}$

54 .

$\frac{3+\mathrm{ii}i}{2+i}+\left(4+3i\right)$

55 .

$\frac{4+i}{i}+\frac{\mathrm{iii}-4i}{1-i}$

56 .

$\frac{3+2i}{1+2i}-\frac{\mathrm{ii}-3i}{3+i}$

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Source: https://openstax.org/books/college-algebra-2e/pages/2-4-complex-numbers

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