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Chapter 6: Problem 24

Describe the subset of the complex plane consisting of the complex numbers \(z\) such that \(z^{3}\) is a real number.

Short Answer

Expert verified
The subset of the complex plane consisting of complex numbers z such that \(z^3\) is a real number is the set of complex numbers with arguments \(\theta = \frac{n\pi}{3}\), where n is an integer (with no restrictions on the magnitude r).

Step by step solution

01

Express the complex number in polar form

For a complex number z, we can express it in polar form as follows: \(z = re^{i\theta}\), where r is the magnitude of z and \(\theta\) is the argument of z.
02

Compute the cube

Compute the cube of the complex number in polar form: \(z^3 = (re^{i\theta})^3 = r^3e^{i3\theta}\).
03

Find conditions for real solutions

For a complex number to be real, its imaginary part must be zero. For this to happen in a polar representation, the exponent of \(e\) must be a multiple of \(\pi\) so that the imaginary part vanishes. So, we must have: \(3\theta = n\pi\), where n is an integer.
04

Solve for the argument of z

If \(3\theta = n\pi\), then \(\theta = \frac{n\pi}{3}\).
05

Interpret the solution

For the complex number z to have its cube real, its argument must be \(\theta = \frac{n\pi}{3}\), where n is an integer. Therefore, the subset of the complex plane consisting of complex numbers z such that \(z^3\) is a real number is the set of complex numbers with arguments \(\theta = \frac{n\pi}{3}\) where n is an integer (with no restrictions on the magnitude r).

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