/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 987 Find the value of \((4-4 i) \cdo... [FREE SOLUTION] | 91影视

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Chapter 31: Problem 987

Find the value of \((4-4 i) \cdot(\sqrt{3}-i)\) in polar form.

Short Answer

Expert verified
The value of \((4-4 i) \cdot(\sqrt{3}-i)\) in polar form is \(8\sqrt{2} \operatorname{cis}(-\frac{5\pi}{12})\).

Step by step solution

01

Convert the given complex numbers to polar form

To convert a complex number to polar form, we need to find its magnitude (r) and its angle (胃). For a complex number in the form of \(a + bi\), \(r = \sqrt{a^2 + b^2}\) and \(\theta = \operatorname{atan2}(b, a)\). For \((4 - 4i)\): \(r_1 = \sqrt{4^2 + (-4)^2} = \sqrt{32} = 4\sqrt{2}\), and \(\theta_1 = \operatorname{atan2}(-4, 4) = -\frac{\pi}{4}\). So, in polar form, \((4-4i) = 4\sqrt{2}\operatorname{cis}(-\frac{\pi}{4})\). For \((\sqrt{3} - i)\): \(r_2 = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{4} = 2\), and \(\theta_2 = \operatorname{atan2}(-1, \sqrt{3}) = -\frac{\pi}{6}\). So, in polar form, \((\sqrt{3}-i) = 2\operatorname{cis}(-\frac{\pi}{6})\).
02

Compute the product of polar forms of complex numbers

When multiplying two complex numbers in polar form, \(r_1 \operatorname{cis}(\theta_1) \cdot r_2 \operatorname{cis}(\theta_2)\), we multiply their magnitudes, \(r_1r_2\), and add their angles, \(\theta_1 + \theta_2\). Thus, the product is: \(r_1 r_2 \operatorname{cis} (\theta_1 + \theta_2) = 4\sqrt{2} \cdot 2 \operatorname{cis}(-\frac{\pi}{4} - \frac{\pi}{6})\). Evaluating it, we get: \(8\sqrt{2} \operatorname{cis}(-\frac{\pi}{4} - \frac{\pi}{6})\).
03

Represent the result in polar form

From step 2, we have the product in polar form as \(8\sqrt{2} \operatorname{cis}(-\frac{\pi}{4} - \frac{\pi}{6})\). Now, we just need to simplify the angle part: \( -\frac{\pi}{4} - \frac{\pi}{6} = -\frac{3\pi + 2\pi}{12} = -\frac{5\pi}{12} \). So the final result in polar form is: \(8\sqrt{2} \operatorname{cis}(-\frac{5\pi}{12})\).

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