Limit of ln(complex)

  • Thread starter MaxManus
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  • #1
MaxManus
277
1

Homework Statement



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


The Attempt at a Solution


I am using taylor where f = [tex]\frac{1-\frac{ix}{z}}{1+\frac{ix}{z}}[/tex]
f'(0) -[tex]\frac{2i}{z}[/tex]

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

Answers and Replies

  • #2
rsa58
85
0
write the equation clearly.
 
  • #3
MaxManus
277
1
I am sorry, but English is not my native language. What does "write the equation clearly" mean?
 
  • #4
36,214
8,196

Homework Statement



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


The Attempt at a Solution


I am using taylor where f = [tex]\frac{1-\frac{ix}{z}}{1+\frac{ix}{z}}[/tex]
f'(0) -[tex]\frac{2i}{z}[/tex]

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
[tex]\lim_{x \to 0} \frac{ln(1 - \frac{ix}{z})}{1 + \frac{ix}{z}}[/tex]

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

You can see my LaTeX script by double-clicking either of the limit expressions above.
 
  • #5
MaxManus
277
1
Thanks, yes I ment the last one
[tex]
\lim_{x \to 0} ln \left(\frac{1 - \frac{ix}{z}}{1 + \frac{ix}{z}}\right)
[/tex]
 
Last edited:
  • #6
36,214
8,196
Is z a constant in your expression? If not, I think you need to use partial derivatives.
 
  • #7
MaxManus
277
1
Yes, it is a constant.
 
  • #9
MaxManus
277
1
Thanks
 

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