Ch. 5 - Complex Numbers, Polar Coordinates and Parametric Equations

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Blitzer 3rd Edition

Ch. 5 - Complex Numbers, Polar Coordinates and Parametric Equations

Problema 68

Capitolo 5, Problema 68

In Exercises 65–68, find all the complex roots. Write roots in polar form with θ in degrees. The complex cube roots of 27(cos 306° + i sin 306°)

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Identify the given complex number in polar form: \(27(\cos 306^\circ + i \sin 306^\circ)\), where the modulus is \(r = 27\) and the argument is \(\theta = 306^\circ\).

Recall that to find the complex cube roots of a complex number in polar form \(r(\cos \theta + i \sin \theta)\), we use De Moivre's Theorem for roots: the \(n\)th roots are given by \(\sqrt[n]{r} \left( \cos \frac{\theta + 360^\circ k}{n} + i \sin \frac{\theta + 360^\circ k}{n} \right)\), where \(k = 0, 1, ..., n-1\).

Calculate the cube root of the modulus: \(\sqrt[3]{27} = 3\).

Find the three arguments for the cube roots by substituting \(n=3\) and \(k=0,1,2\) into the formula for the argument: \(\frac{306^\circ + 360^\circ k}{3}\).

Write each root in polar form as \(3 \left( \cos \alpha_k + i \sin \alpha_k \right)\), where \(\alpha_k\) are the three arguments found in the previous step.

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Complex Numbers in Polar Form

Complex numbers 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 simplifies multiplication, division, and finding roots by working with magnitudes and angles separately.

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Percorso guidato

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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 power n results in r^n (cos nθ + i sin nθ). Conversely, the nth roots can be found by taking the nth root of the magnitude and dividing the angle by n, adding multiples of 360°/n for all roots.

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Percorso guidato

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Powers Of Complex Numbers In Polar Form (DeMoivre's Theorem)

Finding Complex Roots

To find the nth roots of a complex number, calculate the nth root of the magnitude and determine the arguments by dividing the original angle by n and adding k(360°/n) for k = 0, 1, ..., n-1. This yields all distinct roots evenly spaced around the circle.

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Percorso guidato

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Complex Roots

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