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Chapter 1: Problem 7

Find the geometric images of the complex numbers \(z\) such that $$ \left|z+\frac{1}{z}\right|=2 $$

Short Answer

Expert verified
Answer: The geometric images of the complex numbers z that satisfy |z + 1/z| = 2 are points on the unit circle with arguments in the form of 2n蟺 or (2n+1)蟺, where n is any integer.

Step by step solution

01

Rewrite the complex numbers in geometric form

To rewrite the complex numbers in a geometric form, let z = re^(i胃), where r is the modulus (magnitude) and 胃 is the argument (angle) of a complex number. Now, we can rewrite the given equation as follows: $$ \left|re^{i胃} + \frac{1}{re^{i胃}}\right| = 2 $$
02

Simplify the equation and isolate z

Simplify the equation by combining the terms with the same base (in this case, e^(i胃)): $$ \left|r + \frac{1}{r}\right|e^{i胃} = 2 $$ From this equation, we can see that the magnitude of the complex number is $$ \left|r + \frac{1}{r}\right| = 2 $$ Now, let's square both sides to get rid of the modulus sign: $$ \left(r + \frac{1}{r}\right)^2 = 4 $$ Expanding the equation gives: $$ r^2 + 2 + \frac{1}{r^2} = 4 $$ Multiply every term by r^2 to simplify the equation: $$ r^4 + 2r^2 + 1 = 4r^2 $$ Rearrange the equation in a quadratic form: $$ r^4 - 2r^2 + 1 = 0 $$
03

Solve the quadratic equation for r

The equation is a quadratic equation in r^2. Let's solve for r^2 by using the quadratic formula: $$ r^2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ In our case, a = 1, b = -2, and c = 1, so we get: $$ r^2 = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(1)}}{2(1)} $$ Using the formula, we find two possible solutions for r^2: $$ r^2 = 1 $$ This means that r can have two possible values, r = 1 or r = -1. However, since the modulus of a complex number must be non-negative, we can discard the negative value, and we're left with r = 1.
04

Find the possible complex numbers z

Now, as we have found the modulus r = 1, let's find the possible complex numbers z. Since r = 1, the given equation simplifies to: $$ \left|e^{i胃} + \frac{1}{e^{i胃}}\right| = 2 $$ Note that: $$ \frac{1}{e^{i胃}} = e^{-i胃} $$ Thus, the equation becomes: $$ \left|e^{i胃} + e^{-i胃}\right| = 2 $$ Recall the Euler's formula: $$ e^{i胃} = \cos胃 + i\sin胃 $$ So, we can rewrite our equation as: $$ \left|\cos胃 + i\sin胃 + \cos(-胃) + i\sin(-胃)\right| = 2 $$ Using the properties of the sine and cosine functions, we can simplify the equation: $$ \left|2\cos胃\right| = 2 $$ Divide both sides by 2 to get: $$ |\cos胃| = 1 $$
05

Determine the geometric representation of z

We have now found that: $$ |\cos胃| = 1 $$ This equation holds true if and only if 胃 = 2n蟺 or 胃 = (2n+1)蟺, where n is any integer. Since r = 1, the geometric images of the complex numbers z will be points on the unit circle, and their arguments can take any value in the form of 2n蟺 or (2n+1)蟺.

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Most popular questions from this chapter

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

Solve in \(\mathbb{C}\) the equations: (a) \(z^{2}+z+1=0 ;\) (b) \(z^{3}+1=0\).

Let \(n>2\) be an integer. Find the number of solutions to the equation $$ z^{n-1}=i \bar{z} $$

Represent the geometric images of the following complex numbers: $$ \begin{aligned} &z_{1}=3+i ; z_{2}=-4+2 i ; z_{3}=-5-4 i ; z_{4}=5-i ; \\ &z_{5}=1 ; z_{6}=-3 i ; z_{7}=2 i ; z_{8}=-4 \end{aligned} $$

Let \(z_{1}, z_{2}, z_{3}\) be distinct complex numbers such that $$ \left|z_{1}\right|=\left|z_{2}\right|=\left|z_{3}\right|>0 $$ If \(z_{1}+z_{2} z_{3}, z_{2}+z_{1} z_{3}\), and \(z_{3}+z_{1} z_{2}\) are real numbers, prove that \(z_{1} z_{2} z_{3}=1\)
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