# Complex Exponentials

1. Sep 25, 2008

### kevi555

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!

2. Sep 25, 2008

### Philosophaie

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
3. Sep 25, 2008

### kevi555

Please note, I made an error in my first post...it is now correct.

4. Sep 25, 2008

### Philosophaie

This may help:

z=A*cosΘ+i*A*sinΘ

5. Sep 25, 2008

### kevi555

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?

Last edited: Sep 25, 2008
6. Sep 25, 2008

### Philosophaie

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.

7. Sep 25, 2008

### buffordboy23

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. =)