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Precalculus Honors Review 6.5, 9.6 through 9.8
Name: __________________________________ Date: ___________________
6.5- Trigonometric Form of Complex Numbers
οƒ˜ Absolute Value of Complex Numbers
o |π‘Ž + 𝑏𝑖| = βˆšπ‘Ž2 + 𝑏 2
οƒ˜ Trigonometric Form of a Complex Number
o 𝑧 = π‘Ž + 𝑏𝑖
o 𝑧 = π‘Ÿ(cos πœƒ + 𝑖 sin πœƒ)
o π‘Ž = π‘Ÿ cos πœƒ
o 𝑏 = π‘Ÿ sin πœƒ
o π‘Ÿ = βˆšπ‘Ž2 + 𝑏 2
𝑏
o πœƒ = tanβˆ’1 π‘Ž
οƒ˜ Product of Two Complex Numbers
o 𝑧1 = π‘Ÿ1 (cos πœƒ1 + 𝑖 sin πœƒ1 )
o 𝑧2 = π‘Ÿ2 (cos πœƒ2 + 𝑖 sin πœƒ2 )
o 𝑧1 𝑧2 = π‘Ÿ1 π‘Ÿ2 (cos πœƒ1 + πœƒ2 + 𝑖 sin πœƒ1 + πœƒ2 )
οƒ˜ Quotient of Two Complex Numbers
o 𝑧1 = π‘Ÿ1 (cos πœƒ1 + 𝑖 sin πœƒ1 )
o 𝑧2 = π‘Ÿ2 (cos πœƒ2 + 𝑖 sin πœƒ2 )
𝑧
π‘Ÿ
o 𝑧1 = π‘Ÿ1 (cos πœƒ1 βˆ’ πœƒ2 + 𝑖 sin πœƒ1 βˆ’ πœƒ2 ), 𝑧2 β‰  0
2
2
οƒ˜ DeMoivre’s Theorem
o 𝑧 𝑛 = [π‘Ÿ(cos πœƒ + 𝑖 sin πœƒ)]𝑛 = π‘Ÿ 𝑛 (cos π‘›πœƒ + 𝑖 sin π‘›πœƒ)
οƒ˜ nth Root of a Complex Number
o
𝑛
βˆšπ‘Ÿ (cos
πœƒ+2πœ‹π‘˜
𝑛
9.6- Polar Coordinates
οƒ˜ Polar Coordinates
o (π‘Ÿ, πœƒ)
o π‘₯ = π‘Ÿ cos πœƒ
o 𝑦 = π‘Ÿ sin πœƒ
𝑏
o πœƒ = tanβˆ’1 π‘Ž
o π‘Ÿ2 = π‘₯2 + 𝑦2
+ 𝑖 sin
πœƒ+2πœ‹π‘˜
𝑛
) , π‘˜ = 0, 1, 2, … . , 𝑛 βˆ’ 1
Block: _________
9.7- Graphs of Polar Equations
οƒ˜ Testing for Symmetry
πœ‹
o The line πœƒ = 2 : Replace (π‘Ÿ, πœƒ) 𝑏𝑦 (π‘Ÿ, πœ‹ βˆ’ πœƒ) π‘œπ‘Ÿ (βˆ’π‘Ÿ, βˆ’πœƒ)
o The polar axis: Replace (π‘Ÿ, πœƒ) 𝑏𝑦 (π‘Ÿ, βˆ’πœƒ) π‘œπ‘Ÿ (βˆ’π‘Ÿ, πœ‹ βˆ’ πœƒ)
o The pole: Replace (π‘Ÿ, πœƒ) 𝑏𝑦 (π‘Ÿ, πœ‹ + πœƒ) π‘œπ‘Ÿ (βˆ’π‘Ÿ, πœƒ)
οƒ˜ Special Polar Graphs
o Limacons: π‘Ÿ = π‘Ž ± 𝑏 cos πœƒ π‘œπ‘Ÿ π‘Ÿ = π‘Ž ± 𝑏 sin πœƒ , (π‘Ž > 0, 𝑏 > 0)
π‘Ž

Inner Loop if 𝑏 < 1

Cardioid (Heart Shaped): 𝑏 = 1

Dimpled: 1 <

Convex: 𝑏 > 2
π‘Ž
π‘Ž
𝑏
π‘Ž
<2
o Rose Curves: π‘Ÿ = π‘Ž cos π‘›πœƒ π‘œπ‘Ÿ π‘Ÿ = π‘Ž sin π‘›πœƒ
 n petals if n is odd
 2n petals if n is even (𝑛 β‰₯ 2)
o Circles: π‘Ÿ = π‘Ž cos πœƒ π‘œπ‘Ÿ π‘Ÿ = π‘Ž sin πœƒ
o Lemniscates: π‘Ÿ 2 = π‘Ž2 cos 2πœƒ π‘œπ‘Ÿ π‘Ÿ 2 = π‘Ž2 sin 2πœƒ
9.8- Conics
οƒ˜ Equations of Conics
𝑒𝑝
o π‘Ÿ=
π‘œπ‘Ÿ π‘Ÿ =
1±π‘’ cos πœƒ
o Ellipse: 0 < 𝑒 < 1
o Parabola: 𝑒 = 1
o Hyperbola: 𝑒 > 1
𝑒𝑝
1±π‘’ sin πœƒ
, π‘€β„Žπ‘’π‘Ÿπ‘’ 𝑒 > 0
Directions: Write the following in the indicated form.
1. π‘Ÿ = βˆ’2 csc πœƒ in rectangular form
1
2. π‘Ÿ = 2 sin πœƒ in rectangular form
3. 5𝑦 βˆ’ 𝑦 2 = π‘₯ 2 in polar form
Directions: Plot the following points and given an alternate form for the points.
4. A(2, βˆ’
7πœ‹
6
)
πœ‹
5. B(βˆ’3, 6 )
ο€±
πœ‹
6. C(4, 3 )
πœ‹
7. D(βˆ’5, βˆ’ 4 )
Directions: Convert the given points as indicated.
8. Give rectangular coordinates for (βˆ’3, 150°)
9. Give polar coordinates for (2√3, βˆ’2), π‘˜π‘’π‘’π‘ πœƒ 𝑖𝑛 π‘Ÿπ‘Žπ‘‘π‘–π‘Žπ‘›π‘ 
ο€²
ο€³


Directions: Keep all the angles in terms of Ο€ and all other values in simplest radical form where
appropriate. Write the values in the indicated form.
10. Write in trigonometric form: 𝑧1 = βˆ’5𝑖
11. Write in trigonometric form: 𝑧2 = √3 + 3𝑖
12. Calculate 𝑧1 𝑧2 using polar form (use 𝑧1 and 𝑧2 from problems 10 and 11).
13. Convert #12 into rectangular form.
14. Use DeMoivre’s Theorem to find (𝑧2 )3 (use 𝑧2 from problem 11).
15. Express your answer to 14 in rectangular form.
Directions: Keep all the angles in degrees and round all answers to the nearest hundredth.
16. Find the fourth roots of -81 using trigonometric form.
17. Convert your answer to 16 into rectangular form.
18. Use DeMoivre’s Theorem to show that 2 cos 45° is a fifth root of βˆ’16√2 βˆ’ 16√2𝑖.
19. Find the other fifth roots of βˆ’16√2 βˆ’ 16√2𝑖 in trigonometric form.
20. Perform the indicated operation. Leave your answer in trigonometric form.
12(cos 37° + 𝑖 sin 37°)
42(cos 12° + 𝑖 sin 12°)
Directions: Identify the type of graph given the equation. If it is a rose curve, identify the
number of petals. If it is a limacon, identify the specific limacon.
21. π‘Ÿ = 3 cos 2πœƒ
22. π‘Ÿ = 5 βˆ’ 5 sin πœƒ
23. π‘Ÿ 2 = 9 cos 2πœƒ
4
27. π‘Ÿ = 1βˆ’cos πœƒ
3
28. π‘Ÿ = 2βˆ’cos πœƒ
3
29. π‘Ÿ = 2+cos πœƒ
4
24. π‘Ÿ = 3 cos πœƒ
30. π‘Ÿ = 1βˆ’3sin πœƒ
25. π‘Ÿ = 6 sin 2πœƒ
31. π‘Ÿ = 1+2 sin πœƒ
26. π‘Ÿ = 1 + 4 cos πœƒ
3
4
32. π‘Ÿ = 1+sin πœƒ
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