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Chapter 6: Problem 5

Write \(2-2 i\) in polar form.

Short Answer

Expert verified
The polar form of the complex number \(2 - 2i\) is \(\sqrt{8}(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4})\).

Step by step solution

01

Identify the real and imaginary parts

For the complex number \(2-2i\), the real part \(a\) is \(2\), and the imaginary part \(b\) is \(-2\).
02

Calculate the magnitude of the complex number

We will use the magnitude formula \(r = \sqrt{a^2 + b^2}\). In our case, \(a = 2\) and \(b = -2\). So we have: \[r = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8}.\] The magnitude of the complex number is \(\sqrt{8}\).
03

Calculate the angle of the complex number

We will use the angle formula \(\theta = \arctan\left(\frac{b}{a}\right)\). In our case, \(a = 2\) and \(b = -2\). So we have: \[\theta = \arctan\left(\frac{-2}{2}\right) = \arctan(-1).\] Since the complex number is in the third quadrant, we should add \(\pi\) to the angle: \[\theta = \arctan(-1) + \pi = -\frac{\pi}{4} + \pi = \frac{3\pi}{4}.\] The angle of the complex number is \(\frac{3\pi}{4}\).
04

Write the polar form

Now that we have the magnitude and the angle, we can write the polar form of the complex number. The polar form of a complex number is given by: \[z = r(\cos\theta + i\sin\theta)\] In our case, \(r = \sqrt{8}\) and \(\theta = \frac{3\pi}{4}\). So we have: \[2 - 2i = \sqrt{8}(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4}).\] The polar form of the complex number \(2 - 2i\) is \(\sqrt{8}(\cos\frac{3\pi}{4} + i\sin\frac{3\pi}{4})\).

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Most popular questions from this chapter

Show that \(\overline{w-z}=\bar{w}-\bar{z}\) for all complex numbers \(\mathcal{w}\) and \(z\).

Write out a table showing the values of \(i^{n}\) with \(n\) ranging over the integers from 1 to 12 . Describe the pattern that emerges.

Write each expression in the form \(a+b i,\) where \(a\) and \(b\) are real numbers. $$ (2+3 i)^{3} $$

Show that \(\overline{\bar{z}}=z\) for every complex number \(z\).

In Example 9 we found that the angle \(\theta\) equals \(\tan ^{-1} 2-\tan ^{-1} \frac{1}{3}\) and also that \(\theta\) equals \(\frac{\pi}{4}\). Thus $$ \tan ^{-1} 2-\tan ^{-1} \frac{1}{3}=\frac{\pi}{4} $$ (a) Use one of the inverse trigonometric identities from Section 5.2 to show that the equation above can be rewritten as $$ \tan ^{-1} 2+\tan ^{-1} 3=\frac{3 \pi}{4} $$ (b) Explain how adding \(\frac{\pi}{4}\) to both sides of the equation above leads to the beautiful equation $$ \tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi $$.
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