Limit of ln(complex): Find Derivative at x=0

[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?

 

[rsa58]

write the equation clearly.

 

[MaxManus]

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

 

[Mark44]

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
<br /> \lim_{x \to 0} ln \left(\frac{1 - \frac{ix}{z}}{1 + \frac{ix}{z}}\right)<br />

 

[Mark44]

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

 

[MaxManus]

Yes, it is a constant.

 

[Mark44]

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

 

[MaxManus]

Thanks