# Limit of ln(complex)

MaxManus

## Homework Statement

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

## 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?

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rsa58
write the equation clearly.

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

Mentor

## Homework Statement

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

## 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?

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.

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

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

MaxManus
Yes, it is a constant.

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

MaxManus
Thanks