Simple complex algebra problem

[flemmyd]

Homework Statement

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

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.

 

[mighty2000]

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!