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Chapter 6: Problem 44

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

Short Answer

Expert verified
The product of \(z_1 = 1+i\) and \(z_2 = 2+2i\) in polar form is \(4(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2})\).

Step by step solution

01

Convert to Polar Form

Let's start by converting the complex numbers \(z_1=1+i\) and \(z_2=2+2i\) into polar form. The polar form of a complex number \(z=a+bi\) can be found using \(r = |z| = \sqrt{a^2 + b^2}\) as the magnitude and \(\theta = atan2(b, a)\) as the angle. For \(z_1=1+i\), we get \(r_1=\sqrt{1^2 + 1^2} = \sqrt{2}\) and \(\theta_1 = atan2(1, 1) = \pi/4\) radians, giving the polar form as \(z_1 = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})\). Similarly for \(z_2=2+2i\), we have \(r_2=\sqrt{2^2 + 2^2} = 2\sqrt{2}\) and \(\theta_2 = atan2(2, 2) = \pi/4\) radians. Hence, the polar form of \(z_2\) is \(2\sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})\).
02

Multiply Complex Numbers in Polar Form

To multiply two complex numbers in polar form, simply multiply their magnitudes and add their angles. So, \(z_1 \cdot z_2 = (r_1 \cdot r_2)(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))\). Substituting \(r_1 =\sqrt{2}\), \(\theta_1=\pi/4\), \(r_2=2\sqrt{2}\) and \(\theta_2=\pi/4\), the product becomes \(z_1 \cdot z_2 = (\sqrt{2} \cdot 2\sqrt{2})(\cos(\pi/4 + \pi/4) + i \sin(\pi/4 + \pi/4))\) which simplifies to \(4(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2})\).

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Most popular questions from this chapter

Help you prepare for the material covered in the first section of the next chapter. a. Does (4,-1) satisfy \(x+2 y=2 ?\) b. Does (4,-1) satisfy \(x-2 y=6 ?\)

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