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Chapter 6: Problem 42

Find the product of the complex numbers. Leave answers in polar form. $$\begin{aligned} &z_{1}=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\\\ &z_{2}=\cos \frac{\pi}{4}+i \sin \frac{\pi}{4} \end{aligned}$$

Short Answer

Expert verified
The product of the given complex numbers in polar form is \( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \).

Step by step solution

01

Write the complex numbers in polar form

The given complex numbers are already in their polar form. So we can write them as follows: \( z_1=r_1 (\cos\theta_1 + i\sin\theta_1) = (\cos \frac{\pi}{6} + i \sin \frac{\pi}{6}) \) and \( z_2=r_2 (\cos\theta_2 + i\sin\theta_2) = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \). The moduli of the complex numbers, \( r_1 \) and \( r_2\), are both 1, while the arguments are \( \frac{\pi}{6} \) and \( \frac{\pi}{4} \) respectively.
02

Multiply the moduli and add the arguments

For multiplication in polar form, you multiply the moduli (or absolute values) \( r_1 \) and \( r_2 \) and add the arguments \( \theta_1 \) and \( \theta_2 \). Here both \( r_1 \) and \( r_2 \) are 1, and \( \theta_1 = \frac{\pi}{6} \), \( \theta_2 = \frac{\pi}{4} \). Therefore, the modulus of the result is \( r_1 \times r_2 = 1 \times 1 = 1 \) and the argument of the result is \( \theta_1 + \theta_2 = \frac{\pi}{6} + \frac{\pi}{4} = \frac{5\pi}{12} \).
03

Write the result in polar form

The result of the multiplication in polar form is \( r_3 (\cos\theta_3 + i\sin\theta_3) \), where \( r_3 = 1 \) and \( \theta_3 = \frac{5\pi}{12} \). Therefore, the product of the given complex numbers in polar form is \( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \).

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Most popular questions from this chapter

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