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Chapter 6: Problem 98

Explain how to find the product of two complex numbers in polar form.

Short Answer

Expert verified
The product of the two complex numbers in polar form is \(r_1 \cdot r_2(cos(\theta_1 + \theta_2) + i \cdot sin(\theta_1 + \theta_2))\).

Step by step solution

01

Understand the Form

Remember that a complex number in polar form is typically represented as \(r(cos \theta + i \cdot sin \theta)\), where \(r\) is the magnitude and \(\theta\) is the angle in the complex plane. Let's assume we have two complex numbers: \(r_1(cos \theta_1 + i \cdot sin \theta_1)\) and \(r_2(cos \theta_2 + i \cdot sin \theta_2)\).
02

Multiply the Magnitudes

In order to find the product of these two complex numbers, the rule is to first multiply their magnitudes (also known as the modulus or absolute value), so the magnitude of our result will be \(r_1 \cdot r_2\).
03

Add the Angles

Then, the second rule states that we must add up the angles (arguments), in this case adding \(\theta_1\) and \(\theta_2\). So the angle of the result will be \(\theta_1 + \theta_2\).
04

Write the Answer

Finally, the product of our original complex numbers will be represented in polar form by \(r_1 \cdot r_2(cos(\theta_1 + \theta_2) + i \cdot sin(\theta_1 + \theta_2))\).

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Most popular questions from this chapter

Find the angle between \(\mathbf{v}\) and \(\mathbf{w} .\) Round to the nearest tenth of a degree. $$\mathbf{v}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=4 \mathbf{i}-3 \mathbf{j}$$

Find \(\text {pro}_{\mathbf{w}} \mathbf{V}\) Then decompose v into two vectors, \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2},\) where \(\mathbf{v}_{1}\) is parallel to \(\mathbf{w}\) and \(\mathbf{v}_{2}\) is orthogonal to \(\mathbf{w}.\) $$\mathbf{v}=\mathbf{i}+2 \mathbf{j}, \quad \mathbf{w}=3 \mathbf{i}+6 \mathbf{j}$$

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