Not at all. I don't know why you are using "dz" here.

The definition of f'(0) is lim_{z -> 0}(f(z) - f(0))/z = lim_{z -> 0}(conj z)^2/z^2.

It turns out that this limit does not exist

Alternatively you could choose two directions
approaching zero such yielding different limits: (1) along the x-axis,
(2) along the y-axis.

Sorry, I was not really thinking when I was writing this (it's late
after midnight in my time zone). Try to approach along (1+i)t, t real,
to 0. I think this should work.

High school math teacher here, dabbling in a bit of complex analysis
at the moment. I believe I have the following problem answered
correctly, but I'd appreciate some confirmation. The task is to decide
whether or not f(z) is differentiable at z=0. Please pardon any
difficulties with my notation, as I'm not a Usenet regular.

f(z) = (conj z)^2 / z for z != 0, f(z) = 0 for z = 0

Applying the definition of the derivative and conjugate quickly gets
us ("dz" = delta z)...

f'(z) = lim as dz->0 (conj dz)^2 / dz

On Apr 26, 5:11 pm, José Carlos Santos <jcsan...@fc.up.pt> wrote:
Not at all. I don't know why you are using "dz" here.

I'm just going along with the notation in the text I'm browsing at the
moment. It prefers the form

f'(z) = lim_{dz -> 0} (f(z+dz) - f(z)) / dz

where, again, "dz" = delta z, not a differential or something. I just
wasn't sure how to format that here. But the dz itself is not really
different than what you're saying.


The definition of f'(0) is lim_{z -> 0}(f(z) - f(0))/z = lim_{z -> 0}(conj z)^2/z^2.

Right. I made a mistake applying the definition and missed the extra
power of z in the denominator. That is certainly different than what I
had said.


It turns out that this limit does not exist

OK, but I'm unclear as to how one would demonstrate that. Would I use
polar form in a way similar to what I showed originally? If I'm
understanding that correctly (and using your notation), I would get...

lim_{z -> 0}(conj z)^2/z^2 = lim_{r -> 0}e^-4it = e^-4it

where t = theta. Would that be enough to prove that the limit does not
exist, since the value of that limit would depend on the direction
from which you approach zero?

Thanks,
Todd

On Apr 26, 6:07 pm, Jannick Asmus <jannick.n...@web.de> wrote:

Alternatively you could choose two directions
approaching zero such yielding different limits: (1) along the x-axis,
(2) along the y-axis.

The text uses that approach in a couple of examples, and I used it in
a couple of earlier problems. But in this case, due to the squaring of
the terms, wouldn't you get a limit of 1 regardless of whether you
approached along the x-axis or the y-axis? That was why I chose the
polar form approach.

Thanks,
Todd