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January 18th, 2010, 02:23 PM   #1
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different branches different derivatives

In complex analysis, certain types of functions are multiple-valued, such as the logarithmic function.

To find the derivative of any multi-valued function, you need to look at the function on a particular branch.

Entire functions, like the exponential function for example, have the same derivative regardless the branch you consider.

Is there a complex-valued function that has different derivatives depending on the branch you choose?

Can someone give an example of such a function with its derivatives on different branches?

Thanks.
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January 19th, 2010, 01:46 PM   #2
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Re: different branches different derivatives

Quote:
Originally Posted by davedave
In complex analysis, certain types of functions are multiple-valued, such as the logarithmic function.

To find the derivative of any multi-valued function, you need to look at the function on a particular branch.

Entire functions, like the exponential function for example, have the same derivative regardless the branch you consider.

Is there a complex-valued function that has different derivatives depending on the branch you choose?

Can someone give an example of such a function with its derivatives on different branches?

Thanks.
Your question is confusing. A single valued function has one branch, multivalued has multiple branches. So multivalued functions in general will have different derivatives on different branches, while single valued functions have single derivatives.

Example: ?z, with derivative .5/?z. On the two branches the function has opposite sign, and so has the derivative.
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January 20th, 2010, 12:35 AM   #3
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Re: different branches different derivatives

Thanks for your clarification. When I was posting this question, I was really puzzled by the connection among branches, multi-valued functions and derivatives.

May I ask you one more question to further clarify the idea of having different derivatives on different branches of multi-valued functions?

I have a multi-valued function in my mind right now which is the natural logarithmic function, f(z)=In(z) for example.

On the principle branch (-pi, pi], its derivative is 1/z.

How would you find its derivatives on other branches, such as (0, 2pi], (-pi/4, 3*pi/4]?

I mean could you please tell me a way to find the derivatives on other branches on this logarithmic function?

Thanks.
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January 20th, 2010, 01:16 PM   #4
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Re: different branches different derivatives

ln(z) is special in that on the different branches, it differs by a constant value (2?ni), where n is an integer. As a result the derivative (1/z) is the same on all branches.
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January 20th, 2010, 02:36 PM   #5
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Re: different branches different derivatives

Thank you so much for your help, mathman.

I know you are very busy with other posters and I should not keep bothering you with the same topic.

Could you please answer one more question? I promise that this is really the last one.

My next multi-valued function is f(z)=z^(1/3).

In polar form, f(z)=r^(1/3)*e^[(theta+2*k*pi)i/3] where k=0,1,2

So, its derivatives on the three branches where k=0,1,2 are z^(-2/3)/3, z^(-2/3)/3 * e^(pi/3)i, and z^(-2/3)/3 * e^(2*pi/3)i
respectively.

Can you tell me whether the derivatives are correct or not?

I really appreciate your help, mathman.
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January 20th, 2010, 05:53 PM   #6
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You shouldn't have gone from 2*k*pi to k*pi.
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January 20th, 2010, 06:52 PM   #7
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Re:

Quote:
Originally Posted by skipjack
You shouldn't have gone from 2*k*pi to k*pi.
sorry. I don't quite understand what you mean in the quote above. I didn't go from 2*k*pi to k*pi.

Could you please explain it?
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January 21st, 2010, 01:40 PM   #8
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Re: different branches different derivatives

Quote:
Originally Posted by davedave
Thank you so much for your help, mathman.

I know you are very busy with other posters and I should not keep bothering you with the same topic.

Could you please answer one more question? I promise that this is really the last one.

My next multi-valued function is f(z)=z^(1/3).

In polar form, f(z)=r^(1/3)*e^[(theta+2*k*pi)i/3] where k=0,1,2

So, its derivatives on the three branches where k=0,1,2 are z^(-2/3)/3, z^(-2/3)/3 * e^(pi/3)i, and z^(-2/3)/3 * e^(2*pi/3)i
respectively.

Can you tell me whether the derivatives are correct or not?

I really appreciate your help, mathman.
An easy way to answer is to write f(z)=c_k*z^(1/3), where c_k=e^[(2k*pi)i/3] and z is the value on the principal branch. Then f'(z)=[c_k*z^(-2/3)]/3
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