Comments (6)
Thanks. I will check it as soon as possible. :)
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Hi there,
After a closer inspection, there is a missing point in my definition. The value of R cannot be taken arbitrarily. This value must depend on the value we have chosen for M. Otherwise we have the issue you mentioned.
The same applies to the other definitions.
I think I will add that extra information in the definitions to clarify it.
What do you think?
Update: I actually have a mistake. It should be R > max(0,ln M). I will fix it.
from complex-analysis.github.io.
Thank you for responding. I'm a novice (if superannuated) at this, so I may still be misinterpreting something. But I don't think that spelling out the constraints on R fixes the problem. I think your existing M,R rules successfully give rigor to the informal notion of a limit that "as |z| rushes off to infinity (along any path), then |f(z)| must also." But if the path is parallel to the imaginary axis, then |e^z| will be the fixed value e^x, never increasing.
To ask the same question a different way: What is special about the zero in D_1 = {z: Re(z) = 0 }? That is, suppose I pose domain D_3 = { z: Re(z) = 3 }? The limit for my D_3 as z -> infinity is now e^3. Doesn't that contradict D_3's inclusion in domain D? If I wished, I could also "slide" your "D" and D_2 domains over to use Re(z) > 3, Re(z) < 3, and have the same properties for those domains as your originals.
from complex-analysis.github.io.
Hi Wilson,
Thanks for your detailed explanation. I see better what is the issue with my example. I did not consider the case for a path parallel to the imaginary axis, since I missed the point that exp(z) is periodic.
To avoid that issue in my example I need to add an extra restriction to the domain, e.g., considering |Im(z)|< delta < pi/2.
What do you think?
Update: Changed my inequality.
from complex-analysis.github.io.
from complex-analysis.github.io.
Hi,
I think I will change the example.
Thanks again for the feedback.
Happy holidays. :)
from complex-analysis.github.io.
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