Euler's Relationship: Solving Vt=V0ejwt

[geo_alchemist]

May be it's a stupid question but I can't figure it out.

according to Eulers Relationship:
e^j$$\alpha$$=cos$$\alpha$$+jsin$$\alpha$$

on the other hand I have equation:
V_t=V₀cos wt
and it can be rewritten as:
V_t=V₀e^jwt

where V is voltage in AC (see link below)
http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/impcom.html"

In this case cos wt is at the place of cos$$\alpha$$, but what I can't understand is, where did jsin $$\alpha$$ go?

 

[Hurkyl]

I quote, with added emphasis:

The real part of a complex exponential function can be used to represent an AC voltage or current.

 

[geo_alchemist]

that is what I am asking about. and what happens with the imaginary part? And if we can simply neglect it, then why?

 

[malicx]

that is what I am asking about. and what happens with the imaginary part? And if we can simply neglect it, then why?

maybe you would find this thread helpful:
https://www.physicsforums.com/showthread.php?t=342547

 

[Hurkyl]

Two miniature ideas to pay attention to:

1. Re(e^ix) is often more convenient than cos(x)
2. Im(e^ix) may tell you something else that's interesting

 

[geo_alchemist]

After searching through the web, only thing I could conclude and understand is that after several transformation from e^jx=cosx+jsinx I come to:
cosx=(e^ix+e^-ix)/2
but, what I can't understand is the equation:
(e^ix+e^-ix)/2=R{e^ix}
well, the wikipedia says that:

Given the symmetry, we only need to perform the analysis for one right-hand term; the results will be identical for the other. At the end of any calculation, we may return to real-valued sinusoids by further noting that
coswt=R{e^jwt}
In other words, we simply take the real part of the result.

see: http://en.wikipedia.org/wiki/Electrical_impedance" Validity of comples representation.

May be it is simple math, but I can't understand if there is any special meaning of curly brackets.

 

[HallsofIvy]

If z= x+ iy then the real part of z is Rz= x.

It's exactly the same thing as if I say "the x-coordinate of the point (2, 3) is 2".

That's all it is- it's not the { } that is important but the "R".

If $e^{j\omega t}= cos(\omega t)+ j sin(\omega t)$ then the "real part" is $cos(\omega t)$ and the "imaginary part' is $sin(\omega t)$ (notice that both "real part" and "imaginary part" of a complex number are real numbers).

$R(e^{j\omega t}) cos(\omega t)$ and $I(e^{j\omega t}) sin(\omega t)$.