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

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The discussion revolves around the relationship between complex exponentials and their representation in vector form. It starts with the expression z=Ae^{iθ} and derives the differential dz=iz dθ, emphasizing the importance of interpreting i as a directional indicator in the complex plane. Participants explore the polar plot representation, where the real and imaginary components correspond to the x and y axes, respectively. The conversation highlights the significance of understanding how the derivative relates to the rotation of vectors by 90 degrees in the complex plane. Overall, the thread aims to clarify the connection between complex numbers and their geometric interpretation.
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Hi,

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

I know that z = x + iy 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).
 
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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 Acos\theta and Asin\theta on the horizontal and vertical axes, respectively. I don't understand how the z is still in the derivative equation?
 
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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 Acos\theta and Asin\theta 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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