Limit of ln(complex)

1. Mar 4, 2010

MaxManus

1. The problem statement, all variables and given/known data

The limit of ln($$\frac{1-\frac{ix}{z}}{1+\frac{ix}{z}}$$) as x goes to 0

3. The attempt at a solution
I am using taylor where f = $$\frac{1-\frac{ix}{z}}{1+\frac{ix}{z}}$$
f'(0) -$$\frac{2i}{z}$$

Is this correct? Can you take the derivative just as equations without imaginary numbers?

Last edited: Mar 4, 2010
2. Mar 4, 2010

rsa58

write the equation clearly.

3. Mar 4, 2010

MaxManus

I am sorry, but English is not my native language. What does "write the equation clearly" mean?

4. Mar 4, 2010

Staff: Mentor

rsa58 means "write the limit expression clearly."

What you have written looks like
$$\lim_{x \to 0} \frac{ln(1 - \frac{ix}{z})}{1 + \frac{ix}{z}}$$

I think that your limit expression is really this:
$$\lim_{x \to 0} ln \left(\frac{1 - \frac{ix}{z}}{1 + \frac{ix}{z}}\right)$$

You can see my LaTeX script by double-clicking either of the limit expressions above.

5. Mar 4, 2010

MaxManus

Thanks, yes I ment the last one
$$\lim_{x \to 0} ln \left(\frac{1 - \frac{ix}{z}}{1 + \frac{ix}{z}}\right)$$

Last edited: Mar 4, 2010
6. Mar 4, 2010

Staff: Mentor

Is z a constant in your expression? If not, I think you need to use partial derivatives.

7. Mar 4, 2010

MaxManus

Yes, it is a constant.

8. Mar 4, 2010

Staff: Mentor

Then I also get f'(0) = -2i/z

9. Mar 4, 2010

Thanks