TLDR;
This video provides a simplified approach to solving root of complex number problems, a topic frequently encountered in university exams. It covers essential value substitutions for 1, -1, i, and -i, and introduces a step-by-step method for finding roots of complex numbers. The video also demonstrates how to handle complex numbers in the form of a + ib using calculators and trigonometric conversions, and concludes with examples, including finding the continued product of roots.
- Key substitutions for 1, -1, i, and -i.
- Step-by-step method for solving roots of complex numbers.
- Calculator and trigonometric techniques for complex number manipulation.
Introduction [0:00]
The lecture focuses on simplifying the process of finding the roots of complex numbers, a topic commonly found in university exams. The presenter aims to provide an easy-to-follow method, contrasting with the complicated approaches often found in textbooks. The presenter encourages viewers to like and share the video for wider reach.
Essential Value Substitutions [0:38]
The presenter introduces four key value substitutions that simplify complex number root problems:
- For 1, substitute cos(0) + i sin(0).
- For -1, substitute cos(π) + i sin(π).
- For i, substitute cos(π/2) + i sin(π/2).
- For -i, substitute cos(π/2) - i sin(π/2). These substitutions are crucial for solving problems involving roots of complex numbers. Additionally, for expressions like 1 + i, converting to polar form using a calculator is recommended, where r is √2 and θ is π/4. For expressions like 1/2 + i√3/2, convert them into cos and sin form using trigonometric values.
Steps to Find Roots of Complex Numbers [4:26]
The presenter outlines a step-by-step method to find the roots of complex numbers:
- Rewrite the equation in the form x = (complex number)^(1/n).
- Add 2nπ (or 2kπ) to the angle inside the trigonometric functions.
- Take π common from the expression.
- Apply De Moivre's Theorem.
- Substitute values for n (or k) from 0 to n-1 to find all roots.
Example 1: Solving x^6 + 1 = 0 [8:00]
The presenter demonstrates the step-by-step method with the example x^6 + 1 = 0:
- Rewrite as x = (-1)^(1/6).
- Substitute -1 with cos(π) + i sin(π).
- Add 2nπ to the angle: cos(2nπ + π) + i sin(2nπ + π).
- Take π common: cos((2n + 1)π) + i sin((2n + 1)π).
- Apply De Moivre's Theorem: cos((2n + 1)π/6) + i sin((2n + 1)π/6).
- Substitute n = 0, 1, 2, 3, 4, 5 to find the six roots. The presenter notes a pattern where the angles increase by 2π, simplifying the calculation of subsequent roots.
Example 2: Finding All Values and Continued Product [14:43]
The presenter tackles a more complex problem: finding all values of (1/2 + i√3/4)^(1/4) and showing that their continued product is 1.
- Convert 1/2 + i√3/2 into trigonometric form: cos(π/3) + i sin(π/3).
- Rewrite the expression: (cos(π/3) + i sin(π/3) / 4)^(1/4).
- Apply De Moivre's Theorem to the inner expression: cos(π) + i sin(π).
- Rewrite as x = (cos(π) + i sin(π))^(1/4).
- Add 2nπ to the angle: cos(2nπ + π) + i sin(2nπ + π).
- Take π common: cos((2n + 1)π) + i sin((2n + 1)π).
- Apply De Moivre's Theorem: cos((2n + 1)π/4) + i sin((2n + 1)π/4).
- Substitute n = 0, 1, 2, 3 to find the four roots. To find the continued product, the presenter converts the roots to exponential form (e^(iθ)), multiplies them by adding the exponents, and converts the result back to trigonometric form to show that the product equals 1.
Conclusion [24:42]
The presenter encourages viewers to practice similar problems from their textbooks and invites them to join the academy's application for more in-depth learning and university-specific content.
