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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 π has exactly π 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 π = ββ1 be included in what came to be known as the complex number system. With
the introduction of the number π, the square root of any negative number can be represented. For example ββ9 =
β9 β ββ1 = 3π.
Numbers like 3π and πβ2 are called imaginary numbers. Numbers like β2 β π 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 + ππ. For instant ace π₯ = 3 can be written asβ¦ π₯ = 3 + 0π 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) ββ75 b) ββ700 c) ββ2205
Complex Numbers Packet 2
Practice A Simplify each radical completely
1. ββ45
2. ββ112 3. β600
4. 2β300
5. β3ββ180 6. 4ββ28
7. ββ72
8. β40
9. ββ1800
10. β3ββ1715
11. β1β392
12. ββ8232
Complex Numbers Packet 3 Standard: N.CN.1 Objective: I can simplify complex numbers.
The imaginary unit, π is defined to be π = ββ1. Using this definition, it would follow that π2 = β1 because
π2 = π β π = ββ1 β ββ1 = ββ12
= β1 The cycles of i:
π = ββ1
π2 = β1
π3 = βββ1 = βπ
π4 = 1
This is like a clock. After π4 is cycles back on itself and π5 = ββ1 = π 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 Β± ππ, 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) (β2 β π)2
= b) 3π β 3π =
Practice B Use the cycles of π to simplify each of the following
1. ββ2 β ββ2
2. 7 + ββ25 3. (4π)2
4. π2 β π3 β π4
5. (ββ4)3
6. (2π)(5π)2
7. π12
8. π8 β π9 β π10 9. 2π2(3π)3
Complex Numbers Packet 4
10. ββ5 β ββ15
11. ββ81 β ββ25 12. β4ββ3 β ββ3
13. (5π)7
14. 5π7 15. β4 β ββ16
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 + 2π) + (5 β 4π) b) (7 β 5π) β (β2 + 6π)
Practice C Simplify each expression
1. (βπ) + (6π)
2. β4π β 5π 3. (β3π) + (3 + 5π)
4. (β2π) + (5π)
5. (3π) + (4π) 6. (β6 β 2π) + (6 β 5π)
7. β5 + 3π β (4 β 5π)
8. (5 + 6π) + (2 β 7π) 9. (6 β 8π) β (4π) + 7
10. (3 β 4π) β (β5 + 7π)
11. (5 β 6π) + (5π) + (7 + 6π) 12. (7 + 7π) β (β7 β 3π) + (β7 β 8π)
Complex Numbers Packet 5 Quick review of the cycles of π
π = ββ1 π2 = β1
π3 = βββ1 = βπ π4 = 1
Simplify each power of π. Simplify each power of i.
π17 π10
π24 π15
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 + ππ form.
Example 1: Multiply β3(β7 + 6π) Example 2: Multiply 4π(2 + 9π) Example 3: Multiply (β2 + 9π)(β3 β 10π)
Complex Numbers Packet 6
Practice D Simplify each expression
1. β8π β 9π
2. (2π)(4π) 3. 3π(7 β 3π)
4. (β5π)(6 β π)
5. 6(8 β 5π) 6. (3 β 6π)(1 β π)
7. (3 + 7π)(2 + 5π)
8. (1 β 7π)(β6 + 8π) 9. (8 + 6π)2
10. (β6 β 5π)2
11. (3 + π)(3 β π) 12. (5 + 4π)(5 β 4π)
Practice E Simplify each expression. Make sure you take note that this is a mixture of add, subtract, and multiply.
1. (9 + 6π)(2 β π)
2. (2 β 3π) + (9 β 7π) 3. (6 + 7π) β (π β 3)
4. (9π β 7)(2π β 7)
5. (3 β 4π) β (2π + 9) 6. (3 + 4π) + (5 β 7π)
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) 3β15
2β5
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.
Complex Numbers Packet 7 Example: Find the conjugate of each complex number.
b) β3 + 5π c) 6 β 7π
Practice F Find the conjugate of each complex number.
1. 5 β 6π
2. 7 + 6π 3. β7 + 7π
4. β4 β 7π
5. 4 + 5π 6. β1 + π
7. β7 + 4π
8. β1 + 6π 9. 5 β 2π
The reason conjugates are useful is because when you multiply a complex number by itβs 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+π Example 4: Divide
22β7π
4β5π
Practice G Simplify each expression completely. Put your answer in Standard Form.
1. 3
βπ
2. 3+2π
8π
3. β1β5π
β10π
4. 9
3β2π
5. 9π
2+2π
6. β8+7π
β1β10π
Complex Numbers Packet 8
7. β8+7π
β1β10π
8. 9β4π
β3+10π
9. 9β4π
4β5π
10. 2+6π
7β9π
11. 10+9π
β10+4π 12.
7+3π
β4β6π
13. 1β8π
β8β3π
14. 3βπ
β6β7π 15.
5β3π
β6β2π
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 + ππ, π is a real number and is graphed on the real axis. ππ is an imaginary number and is graphed on the imaginary axis. When you graph them, you graph them as vectors. Examples:
a) 3 β 2π
b) β7 + 4π
Complex Numbers Packet 9
Practice H Graph each complex number.
1. 3 + 5π
2. β1 + π
3. 8 + 7π
4. β6 β 3π
5. β2 + 3π
6. 8π
Complex Numbers Packet 10
7. 4 β 7π
8. 9 β 3π
9. β4 + 2π
10. β6 β 3π
Find the complex number of each graph.
11.
12.
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 |π§|, and the modulus of a complex number π§ = π + ππ is given by βπ2 + π2.
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 + 24π
3 + 3β3π β2 β 2π
Complex Numbers Packet 12
Practice I Find the modulus of each complex number.
1. 9 β 7π
2. 6 + 3π
3. β4 + 2π
4. β5 β 8π
5.
6.
7.
8.
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. Letβs review this now.
a) 81π₯2 β 16 b) 64 β 49π₯2 Why does it have to be a subtraction sign? Why couldnβt 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 + 81π¦2
2. 36 + 100π₯2 3. 16π₯2 β 81π¦2
4. π₯2 + 36
5. 16π₯2 + 49 6. π₯2 β 36
7. 225π₯2 + 144π¦2
8. 16π₯2 β 49 9. 81 β 64π₯2
10. 4π₯2 + 25
11. π₯2 + 1 12. π₯2 β 1
Complex Numbers Packet 14
13. 4π₯2 β 25
14. 81 + 64π₯2 15. 9π₯2 + 121
16. 169 β 25π₯2
17. 9π₯2 β 121 18. 169 + 25π₯2
Practice K Evaluate each combined function. If π(π₯) = 3π₯ + 1 and π(π₯) = 3π₯2 β 5π₯ β 2
1. 5π(2)
2. 2π(1) β π(5)
3. π(β2) β π(4)
4. π(β1) + π(3)
5. π(2) β π(4)
6. π(β1) β 3π(1)