Properties of Complex Numbers (phasor notation)

Click For Summary
The discussion focuses on the derivation of the real and imaginary parts of a complex number using phasor notation. The real part is expressed as Real(A) = |A|cos(α), while the imaginary part is Imag(A) = j|A|sin(α). The derivation utilizes Euler's formula, where e^(jα) = cos(α) + jsin(α) and e^(-jα) = cos(α) - jsin(α). This leads to the simplifications resulting in the expressions for the real and imaginary components. The thread emphasizes the mathematical foundations behind these representations in complex number theory.
jeff1evesque
Messages
312
Reaction score
0
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
 
Physics news on Phys.org
jeff1evesque said:
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 = cosα + jsinα
e-jα = cosα - jsinα

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

Similar threads

Replies
2
Views
1K
Replies
19
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 26 ·
Replies
26
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K
Replies
2
Views
2K