Properties of Complex Numbers (phasor notation)

[jeff1evesque]

1. Statement:
The Real Part of a "Complex Number is expressed as the following:
$$Real(A) = \frac{1}{2}(A + A*) = \frac{1}{2}(|A|e^{j\alpha} + |A|e^{-j\alpha}) = \frac{1}{2}|A|(2cos(\alpha)) = |A|cos(\alpha)$$. (#1)

The Imaginary Part of a "Complex Number" is expressed as the following:
$$Imag(A) = \frac{1}{2}(A - A*) = \frac{1}{2}(|A|e^{j\alpha} - |A|e^{-j\alpha}) = \frac{1}{2}|A|(2jsin(\alpha)) = j|A|sin(\alpha)$$. (#2)


2. Questions:
I was just curious how $$\frac{1}{2}|A|(2cos(\alpha))$$ was derived in equation (#1), and how $$\frac{1}{2}|A|(2jsin(\alpha))$$ was derived in equation (#2)?

thanks,


Jeff

 

[tiny-tim]

I was just curious how $$\frac{1}{2}|A|(2cos(\alpha))$$ was derived in equation (#1), and how $$\frac{1}{2}|A|(2jsin(\alpha))$$ was derived in equation (#2)?

e^jα = cosα + jsinα
e^-jα = cosα - jsinα

so e^jα + e^-jα = 2cosα
e^jα - e^-jα = 2jsinα

 

[Mene]

Hello Jeff,

Thank you for your question. In order to understand how the equations (#1) and (#2) were derived, we need to first understand the concept of phasor notation in complex numbers.

Phasor notation is a way of representing complex numbers in the form of magnitude and phase. In this notation, a complex number A is represented as A = |A|e^(jα), where |A| is the magnitude and α is the phase angle.

Now, to derive equation (#1), we use the fact that the real part of a complex number is given by the average of the number and its complex conjugate. In other words, Real(A) = (A + A*)/2. Substituting A = |A|e^(jα) in this equation, we get Real(A) = (|A|e^(jα) + |A|e^(-jα))/2. Using the trigonometric identity e^(jα) + e^(-jα) = 2cos(α), we get Real(A) = |A|cos(α).

Similarly, to derive equation (#2), we use the fact that the imaginary part of a complex number is given by the difference of the number and its complex conjugate. In other words, Imag(A) = (A - A*)/2. Substituting A = |A|e^(jα) in this equation, we get Imag(A) = (|A|e^(jα) - |A|e^(-jα))/2. Using the trigonometric identity e^(jα) - e^(-jα) = 2jsin(α), we get Imag(A) = j|A|sin(α).

I hope this helps to clarify how equations (#1) and (#2) were derived. Let me know if you have any further questions.