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Chapter 6: Problem 72

Find all the complex roots. Write roots in rectangular form. If necessary, round to the nearest tenth. The complex sixth roots of 64

Short Answer

Expert verified
The complex sixth roots of 64 are 卤2, 卤(1 卤 鈭�3i)

Step by step solution

01

Write the Complex Number in Polar Form

First, you need to express the number in its polar form. The complex number 64 can be written in polar form as \( r(cos \: 胃 + i \: sin \: 胃 ) \), where \( r = 鈭�(a^2+b^2) \) is the magnitude and \( 胃 = arctan(b/a) \) is the argument. For the number 64, the polar form is \(64( cos \: 0 + i \: sin \: 0)\) because it lies on the real axis.
02

Apply De Moivre's Theorem to find the Roots

The roots of a complex number \( z^n = r[cos(胃+2k蟺) + i \: sin(胃+2k蟺)] \) are given by \( z_k = \sqrt[n]{r}[cos((胃+2k蟺)/n) + i \: sin((胃+2k蟺)/n)] \), k = 0, 1, 2, ..., n-1. For 64 and n = 6, the six roots are \( z_k = \sqrt[6]{64}(cos((0+2k蟺)/6) + i \: sin((0+2k蟺)/6)) \)
03

Calculate all the Roots from k = 0 to 5

Substitute the values in the formula at step 2, from k = 0 to 5. The roots come out as: \( z_0 = 2(cos(0) + i \: sin(0)) = 2 \), \(z_1 = 2(cos(蟺/3) + i \: sin(蟺/3)) = 1 + 鈭�3i \), \(z_2 = 2(cos(2蟺/3) + i \: sin(2蟺/3)) = -1 + 鈭�3i \), \(z_3 = 2(cos(蟺) + i \: sin(蟺)) = -2 \), \(z_4 = 2(cos(4蟺/3) + i \: sin(4蟺/3)) = -1 - 鈭�3i \), \(z_5 = 2(cos(5蟺/3) + i \: sin(5蟺/3)) = 1 - 鈭�3i \)

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Most popular questions from this chapter

Use the vectors $$\mathbf{u}=a_{1} \mathbf{i}+b_{1} \mathbf{j}, \quad \mathbf{v}=a_{2} \mathbf{i}+b_{2} \mathbf{j}, \quad \text { and } \quad \mathbf{w}=a_{3} \mathbf{i}+b_{3} \mathbf{j},$$ to prove the given property. $$(c \mathbf{u}) \cdot \mathbf{v}=c(\mathbf{u} \cdot \mathbf{v})$$

Let $$\mathbf{u}=-\mathbf{i}+\mathbf{j}, \quad \mathbf{v}=3 \mathbf{i}-2 \mathbf{j}, \quad \text { and } \quad \mathbf{w}=-5 \mathbf{j}$$ Find each specified scalar or vector. $$\operatorname{proj}_{\mathbf{u}}(\mathbf{v}+\mathbf{w})$$

Describe a test for symmetry with respect to the line \(\theta=\frac{\pi}{2}\) in which \(r\) is not replaced.

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

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