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Chapter 1: Problem 27

Find all positive integers \(n\) such that $$ \left(\frac{-1+i \sqrt{3}}{2}\right)^{n}+\left(\frac{-1-i \sqrt{3}}{2}\right)^{n}=2 $$

Short Answer

Expert verified
Based on the step by step solution, provide a short answer to the problem: The positive integer solutions for "n" are of the form \(n = 3k\), where \(k\) is a positive integer. This means that the possible values for "n" are multiples of 3, such as \(3, 6, 9, \dots\).

Step by step solution

01

Rewrite the Complex Numbers as Powers of Roots of Unity

Notice that the two given complex numbers are of the form: $$\frac{-1}{2} \pm \frac{i\sqrt{3}}{2}.$$ These could also be written as: $$\frac{1}{2}(-1 \pm i\sqrt{3}).$$ Since \(-1 + i\sqrt{3}\) and \(-1 - i\sqrt{3}\) are the two other cube roots of \(1\) apart from \(1\), we can write the complex numbers as powers of the cube roots of unity. Specifically, let the cube roots of unity be {1, \(\omega\), \(\omega^2\)}, where: $$\omega = e^{2\pi i/3} = \frac{-1 + i\sqrt{3}}{2},\qquad \omega^2 = e^{4\pi i/3} = \frac{-1 - i\sqrt{3}}{2}.$$ This allows us to rewrite the given equation as: $$\omega^{n}+\omega^{2n}=2.$$
02

Exploit Symmetry and Real Parts of Complex Numbers

Now we know that \(\omega^3 = 1\) and \((\omega^2)^{3}=1= \omega^6\). Using this, we can manipulate the powers of \(\omega\) to simplify the equation further: $$\omega^{n}+\omega^{2n}=\omega^{n}+\omega^{-n}.$$ Since the real part of \(\omega^n\) and \(\omega^{-n}\) are equal, their sum is equal to twice the real part of \(\omega^n\). Thus, the equation becomes: $$2\operatorname{Re}(\omega^{n})=2.$$
03

Find Roots and Solve for "n"

From the previous step, we have: $$\operatorname{Re}(\omega^{n})=1.$$ By Euler's formula, we can write any power of \(\omega\) as: $$\omega^{n}=e^{(2\pi i/3)n}=cos \left(\frac{2\pi n}{3}\right)+i\sin\left(\frac{2\pi n}{3}\right).$$ Therefore, the equation now becomes: $$cos \left(\frac{2\pi n}{3}\right) = 1.$$ From this equation, we find that the cosine value is equal to \(1\) if the angle inside is an integer multiple of \(2\pi\). So, the equation becomes: $$\frac{2\pi n}{3}=2\pi m,$$ where \(m\) is an integer.
04

Find Positive Integer Solutions for "n"

Now we solve for "n": $$n=\frac{3}{2}m.$$ Since "m" is an integer, "n" must be a multiple of \(\frac{3}{2}\) in order to be an integer. The smallest positive integer that is a multiple of \(\frac{3}{2}\) is 3, when \(m = 2\). So the positive integer solutions for "n" are: $$\{2\cdot\frac{3}{2},4\cdot\frac{3}{2},6\cdot\frac{3}{2}, \dots \}= \{3,6,9, \dots\}.$$ Therefore, all positive integer solutions for "n\( are of the form \)n = 3k\( where \)k$ is a positive integer.

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Most popular questions from this chapter

Find the geometric interpretation for the following equalities: (a) \((-5+4 i)+(2-3 i)=-3+i ;\) (b) \((4-i)+(-6+4 i)=-2+3 i\); (c) \((-3-2 i)-(-5+i)=2-3 i\); (d) \((8-i)-(5+3 i)=3-4 i\) (e) \(2(-4+2 i)=-8+4 i\) (f) \(-3(-1+2 i)=3-6 i\).

Let \(x_{1}\) and \(x_{2}\) be the roots of the equation \(x^{2}-x+1=0 .\) Compute the following: (a) \(x_{1}^{2000}+x_{2}^{2000}\); (b) \(x_{1}^{1999}+x_{2}^{1999}\); (c) \(x_{1}^{n}+x_{2}^{n}\), for \(n \in \mathbb{N}\).

Let \(z_{1}, z_{2}, z_{3}\) be complex numbers such that $$ z_{1}+z_{2}+z_{3}=0 \text { and }\left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|=1 \text { . } $$ Prove that $$ z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=0 $$

Let \(z_{0}=(a, b) \in \mathbb{C} .\) Find \(z \in \mathbb{C}\) such that \(z^{2}=z_{0}\).

Consider the complex numbers \(z_{1}, z_{2}, \ldots, z_{n}\) with $$ \left|z_{1}\right|=\left|z_{2}\right|=\cdots=\left|z_{n}\right|=r>0 . $$ Prove that the number $$ E=\frac{\left(z_{1}+z_{2}\right)\left(z_{2}+z_{3}\right) \cdots\left(z_{n-1}+z_{n}\right)\left(z_{n}+z_{1}\right)}{z_{1} \cdot z_{2} \ldots z_{n}} $$ is real.
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