Solving Complex Exponentials: Deduce & Explain Relation w/ Vector Diagram

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Homework Help Overview

The discussion revolves around the relationship between complex exponentials and their representation in vector form, specifically focusing on the expression z = Ae^{i\theta}. Participants are exploring the derivative of this expression and its implications in a vector diagram context.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to deduce the derivative dz = iz dθ and are questioning the roles of i and j in the context of complex numbers versus directional vectors. There is also a focus on how the components of z relate to the polar representation and the implications of the derivative in terms of vector rotation.

Discussion Status

The discussion is active, with participants providing insights and hints to guide understanding. Some are clarifying the relationship between the components of z and the derivative, while others are visualizing the vector representation. There is an acknowledgment of errors and corrections in earlier posts, indicating a collaborative effort to refine understanding.

Contextual Notes

Participants are navigating the complexities of representing complex numbers in both algebraic and geometric forms, with some uncertainty regarding the implications of the derivative in this context. There is a mention of graphing in polar coordinates and the need to clarify the roles of various components in the equations being discussed.

kevi555
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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!
 
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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:
Please note, I made an error in my first post...it is now correct.
 
This may help:

z=A*cosΘ+i*A*sinΘ
 
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:
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.
 
kevi555 said:
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?

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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