Given The Complex Numbers $\[ Z_1 = 30 \left( \cos \frac{8\pi}{5} + I \sin \frac{8\pi}{5} \right) \\] And $\[ Z_2 = 5 \left( \cos \frac{3\pi}{5} + I \sin \frac{3\pi}{5} \right), \\]express The Result Of \[$\frac{z_1}{z_2}\$\]

Introduction

Complex numbers are a fundamental concept in mathematics, and they have numerous applications in various fields, including engineering, physics, and computer science. In this article, we will explore the division of complex numbers using De Moivre's theorem. We will also provide a step-by-step solution to a specific problem involving the division of two complex numbers.

What are Complex Numbers?

A complex number is a number that can be expressed in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit, which satisfies the equation $i^2 = -1$. Complex numbers can be represented graphically on a complex plane, with the real part $a$ on the x-axis and the imaginary part $b$ on the y-axis.

De Moivre's Theorem

De Moivre's theorem is a fundamental concept in complex analysis that states that for any complex number $z = r(\cos \theta + i \sin \theta)$, the $n$th power of $z$ can be expressed as:

$$z^n = r^n(\cos n\theta + i \sin n\theta)$$

This theorem is a powerful tool for simplifying complex number expressions and is widely used in various mathematical and scientific applications.

Expressing Complex Numbers in Polar Form

Complex numbers can be expressed in polar form using the following formula:

$$z = r(\cos \theta + i \sin \theta)$$

where $r$ is the magnitude of the complex number, and $\theta$ is the argument of the complex number. The magnitude $r$ is given by:

$$r = \sqrt{a^2 + b^2}$$

and the argument $\theta$ is given by:

$$\theta = \tan^{-1} \left(\frac{b}{a}\right)$$

The Problem

Given the complex numbers:

$$z_1 = 30 \left( \cos \frac{8\pi}{5} + i \sin \frac{8\pi}{5} \right)$$

and

$$z_2 = 5 \left( \cos \frac{3\pi}{5} + i \sin \frac{3\pi}{5} \right)$$

we need to express the result of $\frac{z_1}{z_2}$.

Solution

To solve this problem, we will use De Moivre's theorem and the formula for dividing complex numbers in polar form.

First, we will express $z_1$ and $z_2$ in polar form:

$$z_1 = 30 \left( \cos \frac{8\pi}{5} + i \sin \frac{8\pi}{5} \right)$$

$$z_2 = 5 \left( \cos \frac{3\pi}{5} + i \sin \frac{3\pi}{5} \right)$$

Now, we will divide $z_1$ by $z_2$ using the formula:

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \left( \cos (\theta_1 - \theta_2) + i \sin (\theta_1 - \theta_2) \right)$$

where $r_1$ and $r_2$ are the magnitudes of $z_1$ and $z_2$, respectively, and $\theta_1$ and $\theta_2$ are the arguments of $z_1$ and $z_2$, respectively.

Plugging in the values, we get:

$$\frac{z_1}{z_2} = \frac{30}{5} \left( \cos \left(\frac{8\pi}{5} - \frac{3\pi}{5}\right) + i \sin \left(\frac{8\pi}{5} - \frac{3\pi}{5}\right) \right)$$

Simplifying, we get:

$$\frac{z_1}{z_2} = 6 \left( \cos \frac{5\pi}{5} + i \sin \frac{5\pi}{5} \right)$$

$$\frac{z_1}{z_2} = 6 \left( \cos \pi + i \sin \pi \right)$$

Using the fact that $\cos \pi = -1$ and $\sin \pi = 0$, we get:

$$\frac{z_1}{z_2} = 6(-1 + 0i)$$

$$\frac{z_1}{z_2} = -6$$

Therefore, the result of $\frac{z_1}{z_2}$ is $-6$.

Conclusion

In this article, we have used De Moivre's theorem and the formula for dividing complex numbers in polar form to solve a specific problem involving the division of two complex numbers. We have shown that the result of $\frac{z_1}{z_2}$ is $-6$. This problem illustrates the power of De Moivre's theorem in simplifying complex number expressions and is a useful tool for solving problems involving complex numbers.

References

• De Moivre, A. (1730). "Miscellanea Analytica de Seriebus et Quadraturis." London: G. Woodfall.
• Euler, L. (1748). "Introductio in Analysin Infinitorum." Lausanne: Marc-Michel Bousquet.
• Rudin, W. (1976). "Principles of Mathematical Analysis." New York: McGraw-Hill.

Further Reading

• Complex Analysis by Serge Lang
• Complex Numbers by Michael Artin
• De Moivre's Theorem by Wolfram MathWorld

Q: What is a complex number?

A: A complex number is a number that can be expressed in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit, which satisfies the equation $i^2 = -1$.

Q: What is De Moivre's theorem?

A: De Moivre's theorem is a fundamental concept in complex analysis that states that for any complex number $z = r(\cos \theta + i \sin \theta)$, the $n$th power of $z$ can be expressed as:

$$z^n = r^n(\cos n\theta + i \sin n\theta)$$

Q: How do I express a complex number in polar form?

A: To express a complex number in polar form, you need to find the magnitude $r$ and the argument $\theta$ of the complex number. The magnitude $r$ is given by:

$$r = \sqrt{a^2 + b^2}$$

and the argument $\theta$ is given by:

$$\theta = \tan^{-1} \left(\frac{b}{a}\right)$$

Q: How do I divide two complex numbers using De Moivre's theorem?

A: To divide two complex numbers using De Moivre's theorem, you need to express both complex numbers in polar form and then use the formula:

$$\frac{z_1}{z_2} = \frac{r_1}{r_2} \left( \cos (\theta_1 - \theta_2) + i \sin (\theta_1 - \theta_2) \right)$$

where $r_1$ and $r_2$ are the magnitudes of $z_1$ and $z_2$, respectively, and $\theta_1$ and $\theta_2$ are the arguments of $z_1$ and $z_2$, respectively.

Q: What is the difference between a complex number and a real number?

A: A complex number is a number that can be expressed in the form $a + bi$, where $a$ and $b$ are real numbers, and $i$ is the imaginary unit. A real number is a number that can be expressed in the form $a$, where $a$ is a real number.

Q: Can I use De Moivre's theorem to find the square root of a complex number?

A: Yes, you can use De Moivre's theorem to find the square root of a complex number. To do this, you need to express the complex number in polar form and then use the formula:

$$\sqrt{z} = \sqrt{r} \left( \cos \frac{\theta}{2} + i \sin \frac{\theta}{2} \right)$$

where $r$ is the magnitude of the complex number, and $\theta$ is the argument of the complex number.

Q: Can I use De Moivre's theorem to find the cube root of a complex number?

A: Yes, you can use De Moivre's theorem to find the cube root of a complex number. To do this, you need to express the complex number in polar form and then use the formula:

$$\sqrt[3]{z} = \sqrt[3]{r} \left( \cos \frac{\theta}{3} + i \sin \frac{\theta}{3} \right)$$

where $r$ is the magnitude of the complex number, and $\theta$ is the argument of the complex number.

Q: What are some common applications of De Moivre's theorem?

A: De Moivre's theorem has numerous applications in various fields, including:

• Electrical engineering: De Moivre's theorem is used to analyze and design electrical circuits, particularly those involving alternating current (AC).
• Signal processing: De Moivre's theorem is used to analyze and process signals, particularly those involving complex numbers.
• Control systems: De Moivre's theorem is used to analyze and design control systems, particularly those involving complex numbers.
• Computer graphics: De Moivre's theorem is used to create 3D models and animations, particularly those involving complex numbers.

Q: Where can I learn more about complex numbers and De Moivre's theorem?

A: There are many resources available to learn more about complex numbers and De Moivre's theorem, including:

• Textbooks: There are many textbooks available on complex analysis and De Moivre's theorem, including "Complex Analysis" by Serge Lang and "Complex Numbers" by Michael Artin.
• Online resources: There are many online resources available, including Wolfram MathWorld and Khan Academy.
• Courses: There are many courses available on complex analysis and De Moivre's theorem, including online courses and in-person courses.

Note: The questions and answers provided are a selection of the many questions and answers that can be asked about complex numbers and De Moivre's theorem.