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Complex Exponentials

  1. Sep 25, 2008 #1
    Hi,

    If [tex]z=Ae^{i\theta}[/tex], deduce that [tex] dz = iz d\theta [/tex], and explain the relation in a vector diagram.

    I know that [tex] z = x + iy [/tex] but I don't know if that's going to help. Any hints or tips would be appreciated! Thanks!
     
  2. jcsd
  3. Sep 25, 2008 #2
    The derivative of z:

    dz=i*A*e^(i*Θ)*dΘ

    You have to ask yourself what are i,j: are they directions (r=xi+yj) or are they complex numbers ( x=a+i*b y=c+j*d).


    Secondly, graphing on a polar plot one direction is for the real (x-axis) and the other for the imaginary (y-axis).
     
    Last edited: Sep 25, 2008
  4. Sep 25, 2008 #3
    Please note, I made an error in my first post...it is now correct.
     
  5. Sep 25, 2008 #4
    This may help:

    z=A*cosΘ+i*A*sinΘ
     
  6. Sep 25, 2008 #5
    Alright, I can picture the [tex]Acos\theta[/tex] and [tex]Asin\theta[/tex] on the horizontal and vertical axes, respectively. I don't understand how the z is still in the derivative equation?
     
    Last edited: Sep 25, 2008
  7. Sep 25, 2008 #6
    dz is the change in z. i in a vector diagram is equal to 90deg, right. So when you take the derivative of a vector you shift 90deg.
     
  8. Sep 25, 2008 #7
    Philosophaie gave you a big hint. Look at the equation in his first post real carefully. What is "z" equal to? You already told us. =)
     
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