Simple complex algebra problem

In summary, we use the complex conjugate property to go from the left term to the right term in the given equation. This involves rewriting the numerator and simplifying the denominator before separating the real and imaginary parts.
  • #1
flemmyd
144
1

Homework Statement



[tex]\frac{\beta}{\alpha + i 2\pi f} = \frac{\alpha \beta}{\alpha^{2}+ 2 \pi f}- i \frac{2 \pi \beta}{\alpha^{2}+ 2 \pi f}[/tex]

Homework Equations



complex conjugate:
(a + ib) * (a - ib) = a^2 + b^2

The Attempt at a Solution


If it matters, this is from a book on Fourier transforms
I'm having trouble seeing this equality, going from the left term to the right. I recognize the first of the left terms (simply taking the complex conjugate of the bottom), but I don't know where the entire second term comes from.

EDIT: Solved. I forgot about the complex conjugate on the top and then just separating the sum in the numerator. and the denominators are the same too... should have been an obvious hint.
 
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  • #2


Hello, thank you for posting your question. It looks like you are trying to understand how to go from the left term to the right term in the given equation. Let's break it down step by step:

1. Starting with the left term, we can see that the numerator is simply the complex conjugate of the denominator. This can be written as:

\frac{\beta}{\alpha + i 2\pi f} = \frac{\beta}{\alpha - i 2\pi f}

2. Now, we can use the complex conjugate property (a + ib) * (a - ib) = a^2 + b^2 to rewrite the numerator as:

\frac{\beta}{\alpha + i 2\pi f} = \frac{\beta (\alpha - i 2\pi f)}{(\alpha + i 2\pi f)(\alpha - i 2\pi f)}

3. Simplifying the denominator, we get:

\frac{\beta}{\alpha + i 2\pi f} = \frac{\beta (\alpha - i 2\pi f)}{\alpha^2 + (2\pi f)^2}

4. Now, expanding the numerator, we get:

\frac{\beta}{\alpha + i 2\pi f} = \frac{\alpha \beta - i 2\pi \beta f}{\alpha^2 + (2\pi f)^2}

5. Finally, separating the real and imaginary parts, we get the right term of the given equation:

\frac{\beta}{\alpha + i 2\pi f} = \frac{\alpha \beta}{\alpha^2 + (2\pi f)^2} - i \frac{2\pi \beta f}{\alpha^2 + (2\pi f)^2}

I hope this helps you understand the equality better. Remember to always use the complex conjugate property when dealing with complex numbers. Keep up the good work in your studies!
 

Related to Simple complex algebra problem

1. What is the difference between simple and complex algebra problems?

Simple algebra problems usually involve only one or two variables and can be solved using basic arithmetic operations. Complex algebra problems, on the other hand, involve multiple variables and require more advanced techniques such as factoring and solving systems of equations.

2. How do I solve a simple complex algebra problem?

To solve a simple complex algebra problem, start by identifying the variables and any given information. Then, use the appropriate algebraic operations to isolate the variable and solve for its value. Remember to follow the order of operations and check your final answer by plugging it back into the original problem.

3. What are some common mistakes to avoid when solving simple complex algebra problems?

Some common mistakes to avoid when solving simple complex algebra problems include forgetting to distribute a negative sign, mixing up the order of operations, and making calculation errors. It is important to double-check your work and be mindful of potential mistakes.

4. How can I check if my answer to a simple complex algebra problem is correct?

You can check your answer by plugging it back into the original problem and seeing if it satisfies the given equations. You can also use a calculator to verify your calculations or ask a friend or teacher to review your work.

5. How can I improve my skills in solving simple complex algebra problems?

Practice is key to improving your skills in solving simple complex algebra problems. Start with simpler problems and gradually work your way up to more complex ones. You can also seek help from a tutor or online resources that provide step-by-step explanations and practice problems.

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