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Name: __________________________________________ Period: __________ Complex Numbers Packet 1 Standard: N.CN.1 Objective: I can simplify complex radicals. Find the equation for each quadratic function: Find the roots (x-intercepts) of both by looking at the graph. Fundamental Theorem of Algebra: Every polynomial of degree has exactly roots. Find the roots by hand. Either use the vertex form method or the quadratic formula. Because the square root of a negative number has no defined values as either a rational or irrational number, Euler proposed that a new number = βˆšβˆ’1 be included in what came to be known as the complex number system. With the introduction of the number , the square root of any negative number can be represented. For example βˆšβˆ’9 = √9 βˆ™ βˆšβˆ’1 = 3 . Numbers like 3 and √2 are called imaginary numbers. Numbers like βˆ’2 βˆ’ that include a real term and an imaginary term are called COMPLEX NUMBERS. Reread the fundamental theorem of algebra. Do you think EVERY quadratic function has two roots? Note: Every root can be written as a complex number in the form of + . For instant ace =3 can be written as… = 3 + 0 What is a complex number? For this lesson we are going to focus on simplifying radicals with a negative under to root. For example: Simplify each of the following a) βˆšβˆ’75 b) βˆšβˆ’700 c) βˆšβˆ’2205

Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

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Page 1: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Name: __________________________________________ Period: __________ Complex Numbers Packet 1 Standard: N.CN.1 Objective: I can simplify complex radicals. Find the equation for each quadratic function:

Find the roots (x-intercepts) of both by looking at the graph. Fundamental Theorem of Algebra: Every polynomial of degree 𝑛 has exactly 𝑛 roots. Find the roots by hand. Either use the vertex form method or the quadratic formula. Because the square root of a negative number has no defined values as either a rational or irrational number, Euler

proposed that a new number 𝑖 = βˆšβˆ’1 be included in what came to be known as the complex number system. With

the introduction of the number 𝑖, the square root of any negative number can be represented. For example βˆšβˆ’9 =

√9 βˆ™ βˆšβˆ’1 = 3𝑖.

Numbers like 3𝑖 and π‘–βˆš2 are called imaginary numbers. Numbers like βˆ’2 βˆ’ 𝑖 that include a real term and an imaginary term are called COMPLEX NUMBERS. Reread the fundamental theorem of algebra. Do you think EVERY quadratic function has two roots? Note: Every root can be written as a complex number in the form of π‘Ž + 𝑏𝑖. For instant ace π‘₯ = 3 can be written as… π‘₯ = 3 + 0𝑖 What is a complex number? For this lesson we are going to focus on simplifying radicals with a negative under to root. For example: Simplify each of the following

a) βˆšβˆ’75 b) βˆšβˆ’700 c) βˆšβˆ’2205

Page 2: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 2

Practice A Simplify each radical completely

1. βˆšβˆ’45

2. βˆšβˆ’112 3. √600

4. 2√300

5. βˆ’3βˆšβˆ’180 6. 4βˆšβˆ’28

7. βˆšβˆ’72

8. √40

9. βˆšβˆ’1800

10. βˆ’3βˆšβˆ’1715

11. βˆ’1√392

12. βˆšβˆ’8232

Page 3: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 3 Standard: N.CN.1 Objective: I can simplify complex numbers.

The imaginary unit, 𝑖 is defined to be 𝑖 = βˆšβˆ’1. Using this definition, it would follow that 𝑖2 = βˆ’1 because

𝑖2 = 𝑖 βˆ™ 𝑖 = βˆšβˆ’1 βˆ™ βˆšβˆ’1 = βˆšβˆ’12

= βˆ’1 The cycles of i:

𝑖 = βˆšβˆ’1

𝑖2 = βˆ’1

𝑖3 = βˆ’βˆšβˆ’1 = βˆ’π‘–

𝑖4 = 1

This is like a clock. After 𝑖4 is cycles back on itself and 𝑖5 = βˆšβˆ’1 = 𝑖 and starts all over. The number system can be extended to include the set of complex numbers. A complex number written in standard form is a number π‘Ž Β± 𝑏𝑖, where a and b are real numbers. If a = 0, then the number is called imaginary. If b = 0 then the number is called real. If a and b β‰  0 then the number is complex. Extending the number system to include the set of complex numbers I what allowed us to take the square root of negative numbers. We can use this cycle to simplify a complex expression. Examples: Simplify each of the following expressions

a) (√2 βˆ™ 𝑖)2

= b) 3𝑖 βˆ™ 3𝑖 =

Practice B Use the cycles of 𝑖 to simplify each of the following

1. βˆšβˆ’2 βˆ™ βˆšβˆ’2

2. 7 + βˆšβˆ’25 3. (4𝑖)2

4. 𝑖2 βˆ™ 𝑖3 βˆ™ 𝑖4

5. (βˆšβˆ’4)3

6. (2𝑖)(5𝑖)2

7. 𝑖12

8. 𝑖8 βˆ™ 𝑖9 βˆ™ 𝑖10 9. 2𝑖2(3𝑖)3

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Complex Numbers Packet 4

10. βˆšβˆ’5 βˆ™ βˆšβˆ’15

11. βˆšβˆ’81 βˆ™ βˆšβˆ’25 12. βˆ’4βˆšβˆ’3 βˆ™ βˆšβˆ’3

13. (5𝑖)7

14. 5𝑖7 15. √4 βˆ™ βˆšβˆ’16

Standard: N.NC.2 Objective: I can add, subtract and multiply complex numbers. You can add and subtract complex numbers. Similar to the set of real numbers, you do so by combining like terms. The real parts are like terms and the imaginary parts create like terms. Examples:

a) (3 + 2𝑖) + (5 βˆ’ 4𝑖) b) (7 βˆ’ 5𝑖) βˆ’ (βˆ’2 + 6𝑖)

Practice C Simplify each expression

1. (βˆ’π‘–) + (6𝑖)

2. βˆ’4𝑖 βˆ’ 5𝑖 3. (βˆ’3𝑖) + (3 + 5𝑖)

4. (βˆ’2𝑖) + (5𝑖)

5. (3𝑖) + (4𝑖) 6. (βˆ’6 βˆ’ 2𝑖) + (6 βˆ’ 5𝑖)

7. βˆ’5 + 3𝑖 βˆ’ (4 βˆ’ 5𝑖)

8. (5 + 6𝑖) + (2 βˆ’ 7𝑖) 9. (6 βˆ’ 8𝑖) βˆ’ (4𝑖) + 7

10. (3 βˆ’ 4𝑖) βˆ’ (βˆ’5 + 7𝑖)

11. (5 βˆ’ 6𝑖) + (5𝑖) + (7 + 6𝑖) 12. (7 + 7𝑖) βˆ’ (βˆ’7 βˆ’ 3𝑖) + (βˆ’7 βˆ’ 8𝑖)

Page 5: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 5 Quick review of the cycles of 𝑖

𝑖 = βˆšβˆ’1 𝑖2 = βˆ’1

𝑖3 = βˆ’βˆšβˆ’1 = βˆ’π‘– 𝑖4 = 1

Simplify each power of 𝑖. Simplify each power of i.

𝑖17 𝑖10

𝑖24 𝑖15

Multiplying Complex Numbers

When multiplying complex numbers you follow the same rules as with Real Numbers. Exponent rules apply

for I the same way they do for variables. Remember to simplify your answer and write in π‘Ž + 𝑏𝑖 form.

Example 1: Multiply βˆ’3(βˆ’7 + 6𝑖) Example 2: Multiply 4𝑖(2 + 9𝑖) Example 3: Multiply (βˆ’2 + 9𝑖)(βˆ’3 βˆ’ 10𝑖)

Page 6: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 6

Practice D Simplify each expression

1. βˆ’8𝑖 βˆ™ 9𝑖

2. (2𝑖)(4𝑖) 3. 3𝑖(7 βˆ’ 3𝑖)

4. (βˆ’5𝑖)(6 βˆ’ 𝑖)

5. 6(8 βˆ’ 5𝑖) 6. (3 βˆ’ 6𝑖)(1 βˆ’ 𝑖)

7. (3 + 7𝑖)(2 + 5𝑖)

8. (1 βˆ’ 7𝑖)(βˆ’6 + 8𝑖) 9. (8 + 6𝑖)2

10. (βˆ’6 βˆ’ 5𝑖)2

11. (3 + 𝑖)(3 βˆ’ 𝑖) 12. (5 + 4𝑖)(5 βˆ’ 4𝑖)

Practice E Simplify each expression. Make sure you take note that this is a mixture of add, subtract, and multiply.

1. (9 + 6𝑖)(2 βˆ’ 𝑖)

2. (2 βˆ’ 3𝑖) + (9 βˆ’ 7𝑖) 3. (6 + 7𝑖) βˆ’ (𝑖 βˆ’ 3)

4. (9𝑖 βˆ’ 7)(2𝑖 βˆ’ 7)

5. (3 βˆ’ 4𝑖) βˆ’ (2𝑖 + 9) 6. (3 + 4𝑖) + (5 βˆ’ 7𝑖)

Standard: N.CN.3 Objective I can find the conjugate of a complex number and use it to divide complex numbers. Remember when we were dividing radicals you cannot have a radical in the denominator. To get rid of the radical in the denominator you have to multiply the numerator and the denominator by the radical in the denominator. Examples:

a) 3√15

2√5

Dividing complex numbers is similar. To completely simplify you cannot have an imagery number in the denominator. To get rid of the imaginary number in the denominator you have to multiply by the conjugate of the denominator. What is the conjugate of a complex number? The conjugate of a complex number is the same complex number but opposite sign in between.

Page 7: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 7 Example: Find the conjugate of each complex number.

b) βˆ’3 + 5𝑖 c) 6 βˆ’ 7𝑖

Practice F Find the conjugate of each complex number.

1. 5 βˆ’ 6𝑖

2. 7 + 6𝑖 3. βˆ’7 + 7𝑖

4. βˆ’4 βˆ’ 7𝑖

5. 4 + 5𝑖 6. βˆ’1 + 𝑖

7. βˆ’7 + 4𝑖

8. βˆ’1 + 6𝑖 9. 5 βˆ’ 2𝑖

The reason conjugates are useful is because when you multiply a complex number by it’s conjugate, it ALWAYS a positive real number. To divide complex numbers, find the conjugate of the denominator, multiply the numerator and denominator by

that conjugate, and simplify.

Example 3: Divide 10

2+𝑖 Example 4: Divide

22βˆ’7𝑖

4βˆ’5𝑖

Practice G Simplify each expression completely. Put your answer in Standard Form.

1. 3

βˆ’π‘–

2. 3+2𝑖

8𝑖

3. βˆ’1βˆ’5𝑖

βˆ’10𝑖

4. 9

3βˆ’2𝑖

5. 9𝑖

2+2𝑖

6. βˆ’8+7𝑖

βˆ’1βˆ’10𝑖

Page 8: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 8

7. βˆ’8+7𝑖

βˆ’1βˆ’10𝑖

8. 9βˆ’4𝑖

βˆ’3+10𝑖

9. 9βˆ’4𝑖

4βˆ’5𝑖

10. 2+6𝑖

7βˆ’9𝑖

11. 10+9𝑖

βˆ’10+4𝑖 12.

7+3𝑖

βˆ’4βˆ’6𝑖

13. 1βˆ’8𝑖

βˆ’8βˆ’3𝑖

14. 3βˆ’π‘–

βˆ’6βˆ’7𝑖 15.

5βˆ’3𝑖

βˆ’6βˆ’2𝑖

16.

Standard: N.CN.3 Objective: I can graph a complex number and find the modulus. When graphing linear lines, ordered pairs, quadratic function etc. you graph them on the x and y plane. When graphing a complex number you have to have a complex plane. The complex plane is similar to the x and y plane except that the x-axis is now the real axis and the y-axis is the imaginary axis. In a complex number π‘Ž + 𝑏𝑖, π‘Ž is a real number and is graphed on the real axis. 𝑏𝑖 is an imaginary number and is graphed on the imaginary axis. When you graph them, you graph them as vectors. Examples:

a) 3 βˆ’ 2𝑖

b) βˆ’7 + 4𝑖

Page 9: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 9

Practice H Graph each complex number.

1. 3 + 5𝑖

2. βˆ’1 + 𝑖

3. 8 + 7𝑖

4. βˆ’6 βˆ’ 3𝑖

5. βˆ’2 + 3𝑖

6. 8𝑖

Page 10: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 10

7. 4 βˆ’ 7𝑖

8. 9 βˆ’ 3𝑖

9. βˆ’4 + 2𝑖

10. βˆ’6 βˆ’ 3𝑖

Find the complex number of each graph.

11.

12.

Page 11: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 11

13.

14.

15.

16.

Modulus of a complex number is the distance of the complex number from the origin in a complex plane. The

modulus is denoted by |𝑧|, and the modulus of a complex number 𝑧 = π‘Ž + 𝑏𝑖 is given by βˆšπ‘Ž2 + π‘Ž2.

Notice that the modulus of a complex number is always a real number and in fact it will never be negative since

square roots always return a positive number or zero.

Find the modulus:

18 + 24𝑖

3 + 3√3𝑖 βˆ’2 βˆ’ 2𝑖

Page 12: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 12

Practice I Find the modulus of each complex number.

1. 9 βˆ’ 7𝑖

2. 6 + 3𝑖

3. βˆ’4 + 2𝑖

4. βˆ’5 βˆ’ 8𝑖

5.

6.

7.

8.

Page 13: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 13 Standard: N.RN.1 Objective: I can factor the sum of squares. Earlier this year we learned how to factor difference of squares. Let’s review this now.

a) 81π‘₯2 βˆ’ 16 b) 64 βˆ’ 49π‘₯2 Why does it have to be a subtraction sign? Why couldn’t we factor if it was an addition sign? Because we have learned about imaginary numbers, we can now factor the sum of squares.

Steps Example

1. Rewrite the equation with a subtraction sign. To rewrite it with a subtraction sign you have to change the second term to a negative. So you are subtracting a negative.

81π‘₯2 + 16

2. Take the square root of the first term

3. Take the square root of the second term. Remember that the square root of a negative number is imaginary.

4. Two sets of parenthesis with the square root of the first and second term. One parenthesis has a plus sign, one has a minus.

Examples: a) 9π‘₯2 + 64 b) π‘₯2 + 225

Practice J Factor each completely

1. 16π‘₯2 + 81𝑦2

2. 36 + 100π‘₯2 3. 16π‘₯2 βˆ’ 81𝑦2

4. π‘₯2 + 36

5. 16π‘₯2 + 49 6. π‘₯2 βˆ’ 36

7. 225π‘₯2 + 144𝑦2

8. 16π‘₯2 βˆ’ 49 9. 81 βˆ’ 64π‘₯2

10. 4π‘₯2 + 25

11. π‘₯2 + 1 12. π‘₯2 βˆ’ 1

Page 14: Name: Period: Complex Numbers Packet 1 Standard: N.CN.1 ...€¦ · Complex Numbers Packet 5 Quick review of the cycles of 𝑖 𝑖=βˆšβˆ’1 𝑖2=βˆ’1 𝑖3=βˆ’βˆšβˆ’1=βˆ’π‘– 𝑖4=1

Complex Numbers Packet 14

13. 4π‘₯2 βˆ’ 25

14. 81 + 64π‘₯2 15. 9π‘₯2 + 121

16. 169 βˆ’ 25π‘₯2

17. 9π‘₯2 βˆ’ 121 18. 169 + 25π‘₯2

Practice K Evaluate each combined function. If 𝑓(π‘₯) = 3π‘₯ + 1 and 𝑔(π‘₯) = 3π‘₯2 βˆ’ 5π‘₯ βˆ’ 2

1. 5𝑓(2)

2. 2𝑓(1) βˆ’ 𝑔(5)

3. 𝑓(βˆ’2) βˆ’ 𝑔(4)

4. 𝑔(βˆ’1) + 𝑓(3)

5. 𝑓(2) βˆ™ 𝑔(4)

6. 𝑓(βˆ’1) βˆ™ 3𝑔(1)