Textbook Question
In Exercises 32–35, find all the complex roots. Write roots in rectangular form. The complex fourth roots of 16 (cos 2π/3 + i sin 2π/3)
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11. Graphing Complex Numbers
Powers of Complex Numbers (DeMoivre's Theorem)
3:39 minutesProblem 7
Textbook Question
Perform the indicated operation. Leave answers in polar form. [2(cos 10° + i sin 10°)]⁵
Verified step by step guidance1
Recognize that the expression is in polar form, where the complex number is given by \(r(\cos \theta + i \sin \theta)\) with \(r = 2\) and \(\theta = 10^\circ\).
Recall De Moivre's Theorem, which states that for a complex number in polar form, raising it to the power \(n\) results in \(r^n \left( \cos(n\theta) + i \sin(n\theta) \right)\).
Apply De Moivre's Theorem to the given expression: raise the magnitude to the fifth power, \(r^5 = 2^5\), and multiply the angle by 5, \(5 \times 10^\circ\).
Write the resulting expression as \(2^5 \left( \cos(50^\circ) + i \sin(50^\circ) \right)\), which is the polar form of the complex number raised to the fifth power.
Leave the answer in this polar form without converting to rectangular form or calculating the numerical values.
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3mKey Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Polar Form of Complex Numbers
A complex number can be expressed in polar form as r(cos θ + i sin θ), where r is the magnitude and θ is the argument (angle). This form is useful for multiplication and exponentiation because it separates the magnitude and angle, simplifying calculations involving powers and roots.
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Complex Numbers In Polar Form
De Moivre's Theorem
De Moivre's Theorem states that for a complex number in polar form, raising it to the nth power results in r^n [cos(nθ) + i sin(nθ)]. This theorem allows easy computation of powers of complex numbers by raising the magnitude to the power and multiplying the angle by n.
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Exponentiation of Complex Numbers in Polar Form
To raise a complex number in polar form to a power, apply De Moivre's Theorem by raising the modulus to the power and multiplying the argument by the exponent. This process yields the result in polar form, which is often preferred for clarity and further operations.
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