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chapter5.3r
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à 5.3ïDe Moivre's Formula
äïPlease find the product of the given complex numbers in
êêtrigonometric form.
â
#èGiven z¬ = 3∙(cos 24° + i∙sin 24°) and z½ = 2∙(cos 10° + i∙sin 10°),
#ê z¬∙z½ = 3∙(cos 24° + i∙sin 24°) ∙ 2∙(cos 10° + i∙sin 10°)
êêêè = 6∙(cos 34° + i∙sin 34°).
éSïTwo complex numbers in trigonometric form can be multiplied by
#just multiplying the r values and adding the angles.ïLet z¬ = x¬ + i∙y¬
#and z½ = x½ + i∙y½ be expressed in trigonometric form as r¬(cos Θ +
#i∙sin Θ) and r½(cos Θ½ + i∙sin Θ½).ïThen the product of z¬ and z½ can
#be expressed as z¬∙z½ = r¬(cos Θ¬ + i∙sin Θ) ∙ r½(cos Θ½ + i∙sin Θ½).
If you multiply this out and use the cos and sin of the sum of two an-
#gles formulas, the product simplifies to r¬∙r½(cos (Θ¬ + Θ½) +
#i∙sin (Θ¬ + Θ½)).ïThus, in trigonometric form, the product of two com-
#plex numbers is z¬∙z½ = r¬∙r½(cos (Θ¬ + Θ½) + i∙sin (Θ¬ + Θ½)).
#è In the example, z¬ = 3∙(cos 24° + i∙sin 24°) and z½ = 2∙(cos 10° +
#i∙sin 24°).ïThe product of z¬ and z½ is 6∙(cos 34° + i∙sin 34°).
1ë Find the product of the two complex numbers
#ëz¬ = 4∙(cos 30° + i∙sin 30°) and z½ = 3∙(cos 60° + i∙sin 60°).
è A)ï12∙(cos 30° + i∙sin 30°)ê B)ï12∙(cos 90° + i∙sin 90°)
è C)ï12∙(cos 60° + i∙sin 60°)ê D)ïå of ç
ü
#êêêêëz¬∙z½
êê= 4∙3∙(cos (30° + 60°) + i∙sin (30° + 60°))
êêêï= 12∙(cos 90° + i∙sin 90°)
Ç B
2ë Find the product of the two complex numbers
#ëz¬ = 3∙(cos 5π/6 + i∙sin 5π/6) and z½ = 5∙(cos π/2 + i∙sin π/2).
è A)ï8∙(cos 7π/6 + i∙sin 7π/6)êB)ï12∙(cos π/6 + i∙sin π/6)
è C)ï15∙(cos 4π/3 + i∙sin 4π/3)ë D)ïå of ç
ü
#êêêêëz¬∙z½
êê= 3∙5∙(cos (7π/6 + π/2) + i∙sin (7π/6 + π/2))
êêêï= 15∙(cos 4π/3 + i∙sin 4π/3)
Ç C
3êFind the product of the two complex numbers
#ëz¬ = 2∙(cos 4π/3 + i∙sin 4π/3) and z½ = 5∙(cos 3π/2 + i∙sin 3π/2).
Express answer in rectangular form.
è A)ï-5√3 + 5iêêêïB)ï-4 + 5i
è C)ï2 + √3iêêêè D)ïå of ç
ü
#êêêêëz¬∙z½
êê= 2∙5∙(cos (4π/3 + 3π/2) + i∙sin (4π/3 + 3π/2))
êêêï= 10∙(cos 17π/6 + i∙sin 17π/6)
êêêè = -5√3 + 5i
Ç A
4ë Find the product of (1 + √3∙i) and (1 - i).
êê Express answer in trigonometric form.
è A)ï8∙(cos 365° + i∙sin 365°)êB)ï2∙(cos -45° + i∙sin -45°)
è C)ï2∙√2∙(cos 15° + i∙sin 15°)ë D)ïå of ç
ü
êêêë (1 + √3∙i)∙(1 - i)
êê= 2∙√2∙(cos (60° + 315°) + i∙sin (60° + 315°))
êêêï= 2∙√2∙(cos 375° + i∙sin 375°)
êêêè= 2∙√2∙(cos 15° + i∙sin 15°)
Ç C
äïPlease divide the complex numbers in trigonometric form.
â
16∙(cos 83° + i∙sin 83°)
# ────────────────────────ï=ï(16/8)∙(cos(83° - 23°) + i∙sin(83° - 23°))
ï8∙(cos 23° + i∙sin 23°)
êêêë =ï2∙(cos 60° + i∙sin 60°), trig. form
êêêë =ï1 + √3∙i, rectangular form
éSïTo divide two complex numbers in trigonometric form, you can
just divide the r values and subtract the value of Θ in the denominator
#from the value of Θ in the numerator.ïLet z¬ = r¬(cos Θ¬ + i∙sin Θ¬)
#and z½ = r½(cos Θ½ + i∙sin Θ½).ïThen, z¬/z½ = r¬/r½ ∙ (cos (Θ¬ - Θ½) +
#i∙sin (Θ¬ - Θ½)).ïThis can be proved in a way similar to the product
rule.
è In the example, 16∙(cos 83° + i∙sin 83°) divided by 8∙(cos 23° +
i∙sin 23°) is equal toï(16/8)∙(cos (83° - 23°) + i∙sin (83° - 23°)).
This in turn is equal to 2∙(cos 60° + i∙sin 60°).ïThis is the answer
in trigonometric form.ïIt can also be written in rectangular form as
1 + √3∙i.
5ë Find the quotient of the two complex numbers
#ëz¬ = 4∙(cos 30° + i∙sin 30°) and z½ = 3∙(cos 60° + i∙sin 60°).
è A)ï1.33∙(cos 330° + i∙sin 330°)ëB)ï1/2∙(cos 90° + i∙sin 90°)
è C)ï1.2∙(cos 60° + i∙sin 60°)ê D)ïå of ç
ü
#êêêêëz¬/z½
êê= 4/3∙(cos (30° - 60°) + i∙sin (30° - 60°))
êêêï≈ 1.33∙(cos -30° + i∙sin -30°)
êêêï= 1.33∙(cos 330° + i∙sin 330°)
Ç A
6ë Find the quotient of the two complex numbers
#ëz¬ = 3∙(cos 5π/6 + i∙sin 5π/6) and z½ = 5∙(cos π/2 + i∙sin π/2).
è A)ï.3∙(cos 7π/6 + i∙sin 7π/6)êB)ï.6∙(cos π/3 + i∙sin π/3)
è C)ï1.5∙(cos 4π/3 + i∙sin 4π/3)ë D)ïå of ç
ü
#êêêêëz¬/z½
êê= 3/5∙(cos (7π/6 - π/2) + i∙sin (7π/6 - π/2))
êêêï= .6∙(cos π/3 + i∙sin π/3)
Ç B
7êFind the quotient of the two complex numbers
#ëz¬ = 2∙(cos 4π/3 + i∙sin 4π/3) and z½ = 5∙(cos 3π/2 + i∙sin 3π/2).
Express answer in rectangular form.
è A)ï-.82 + .5iêêê B)ï-.4 + 5i
è C)ï.346 - .2iêêê D)ïå of ç
ü
#êêêêëz¬/z½
êê= 2/5∙(cos (4π/3 - 3π/2) + i∙sin (4π/3 - 3π/2))
êêêï= .4∙(cos -π/6 + i∙sin -π/6)
êêêè ≈ .346 - .2i
Ç C
8ë Find the quotient of (1 + √3∙i) and (1 - i).
êê Express answer in trigonometric form.
è A)ï8∙(cos 315° + i∙sin 315°)êB)ï√2∙(cos -45° + i∙sin -45°)
è C)ï√2∙(cos 105° + i∙sin 105°)ë D)ïå of ç
ü
êêêë (1 + √3∙i)/(1 - i)
êê= 2/√2∙(cos (60° - 315°) + i∙sin (60° - 315°))
êêêï= 2/√2∙(cos -255° + i∙sin -255°)
êêêè= √2∙(cos 105° + i∙sin 105°)
Ç C
äïPlease use De Moivre's Formula to find powers or roots of
êêcomplex numbers.
â
#êè(-√3 + i)Ä = [2∙(cos 150° + i∙sin 150°)]Ä
#êêê= 2Ä∙(cos 3∙150° + i∙sin 3∙150°)
êêê= 8∙(cos 450° + i∙sin 450°)
êêê= 8∙(cos 90° + i∙sin 90°)
êêê= 8i
éSïTo find the square of a complex number in trigonometric form,
#we can use the product rule.ï(r∙(cos Θ + i∙sin Θ))ì = r∙r∙(cos (Θ + Θ)+
#i∙sin (Θ + Θ)) = rì∙(cos 2Θ + i∙sin 2Θ).ïIf the power is greater than
2, then repeated application of the product rule will give us the power
rule or De Moivre's Formula.ï(r∙(cos Θ + i∙sin Θ))ⁿ = rⁿ(cos nΘ +
i∙sin nΘ).
#è In the example, (-√3 + i)Ä = [2∙(cos 150° + i∙sin 150°)]Ä
#êêêêè= 2Ä∙(cos 450° + i∙sin 450°)
êêêêè= 8∙(cos 90° + i∙sin 90°), trig. form
êêêêè= 8i, rectangular form
è In addition to powers of complex numbers, De Moivre's Formula allows
us to find roots of complex numbers.ïIn the past, we have found square
roots and higher roots of real numbers.ïFor example, we know that the
cube root of -1 is -1.ïIt is possible, however, to find two additional
cube roots of -1 by using De Moivre's Formula.ïIn fact, we can find the
roots of any number, real or complex, by using the following formula.
#[r∙(cosΘ + i∙sinΘ)]î¼ⁿ = rî¼ⁿ∙[cos(Θ + 360°∙k)/n + i∙sin(Θ + 360°∙k)/n]
In this formula, k can assume the values k = 0, 1, 2,...,n-1.
è To find the cube roots of -1, we can apply this formula, but first
you should express the number in trigonometric form.
êêë -1 = 1∙(cos 180° + i∙sin 180°)
#Formula.ï(-1)î¼Ä = 1î¼Ä[cos(180° + 360°∙k)/3 + i∙sin(180° + 360°∙k)/3]
#When k = 0, (-1)î¼Ä = 1∙(cos 60° + i∙sin 60°)è= 1/2 + √3i/2
#When k = 1, (-1)î¼Ä = 1∙(cos 180° + i∙sin 180°) = -1
#When k = 2, (-1)î¼Ä = 1∙(cos 300° + i∙sin 300°) = 1/2 - √3i/2
Thus, the three cube roots of -1 are -1, 1/2 + √3i/2, and 1/2 - √3i/2.
It is interesting to note that ç three roots are equally spaced on
the unit circle.ïIn general, the roots of a complex number are equally
spaced on a circle of radius r centered at the origin, where r is the
modulus of the complex number.
# 9ê Find (1 - √3∙i)Å in trigonometric form.
ë A)ï8∙(cos 60° + i∙sin 60°)êB)ï16∙(cos 240° + i∙sin 240°)
ë C)ï2∙(cos 15° + i∙sin 15°)êD)ïå of ç
ü
#êêêêï(1 - √3∙i)Å
#êêêï= [2∙(cos 60° + i∙sin 60°)]Å
#êêêï= 2Å(cos 240° + i∙sin 240°)
êêêï= 16∙(cos 240° + i∙sin 240°)
Ç B
# 10êFind (-1 - i)ì in rectangular form.
ë A)ï2iêêêêB)ï-1 + i
ë C)ï1 - iêêêè D)ïå of ç
ü
#êêêêï(-1 - i)ì
#êêêï= [√2∙(cos 225° + i∙sin 225°)]ì
êêêï= 2∙(cos 450° + i∙sin 450°)
êêêï= 2∙(cos 90° + i∙sin 90°)
êêêï= 2i
Ç A
# 11êFind (3 + 7i)Ä in trigonometric form radian measure.
êêèA)ï441.75∙(cos 3.498 + i∙sin 3.498)
êêèB)ï21.6∙(cos 2.83 + i∙sin 2.83)
êêèC)ï21.6∙(cos 3.5 + i∙sin 3.5)
êêèD)ïå of ç
ü
#êêêêï(3 + 7i)Ä
#êêêï= [7.616∙(cos 1.166 + i∙sin 1.166)]Ä
êêêï= 441.75∙(cos 3.498 + i∙sin 3.498)
Ç A
12êêFind the cube roots of 1.
ë A)ï1,ï1 - i,ï-1 - iêêëB)ï1,ï-1,ïi
ë C)ï1,ï-1/2 + √3i/2,ï-1/2 - √3i/2ë D)ïå of ç
üêêè Find the cube roots of 1.
Express in trig. form.ë1 = 1∙(cos 0° + i∙sin 0°)
#Apply formulaë 1î¼Ä = 1î¼Ä∙[cos(0° +360°k)/3 + i∙sin(0° + 360°∙k)/3]
#For k = 0êè1î¼Ä = 1∙(cos 0° + i∙sin 0°)ê= 1
#For k = 1êè1î¼Ä = 1∙(cos 120° + i∙sin 120°)è= -1/2 + √3i/2
#For k = 2êè1î¼Ä = 1∙(cos 240° + i∙sin 240°)è= -1/2 - √3i/2
Thus, the cube roots of 1 are 1,ï-1/2 + √3i/2,ï-1/2 - √3i/2.
Ç C