hi

@Liad hellow

:P thanks for the answer

i think it should be $f(B(a,r_1) - \{a\})$

mm right

also, $g\circ f$ , im not sure if it is continuous at $a$ , it could not even be defined there

what do you think?>

(for example $e \ ^ {1/z}$)

wait , you assume the limit exists so it is a removable singualarity so we can make $g \circ f$ analytic.

but you assume that $g \circ f$ doesn't have an essential singularity

it is either a removable singularity or a pole;

right so if you assume c is finite then it is removable

sure

but i think my argument works for c not finite too

I see Daniel just posted something in the question.. we can confer with him

yea , what he wrote (the second comment) is exactly what i needed

:) alright! i hope my answer isn't wrong

@Mariah did you see his comment ?

im having a bit trouble with the case it is a pole

yup

just edited my answer

why is the pole case giving you trouble?

you only need to be disjoint from a neighborhood of zero

if we assume $g \circ f = sum_{-m} ^ \{\infty\} a_n (z-a) \ ^ n$

then $g \circ f \ge |a_{-m}| /r \ ^ m$ ?

mm why?

maybe I mean but not sure

i see you accepted, so is it understood?

yea :) but i did it in my way as i wrote in the question

i just want to adjust the case for a pole

thats better!

not sure why it bugs me :/

daniel said $g \circ f >0$ in some nbhd

i can understand why you feel that way

but if we couldn't find an open set around $a$ whose image under g\circ f is disjoint from some open set around zero, we couldn't have the limit be infinity

i see, $lim_{z\to a} g \circ f(z) = \infty$

so by definition we have what you said

thanks :)

of course

btw, do you learn from ash&novinger ?

@Mariah why does $S_1$ finite?

nvm got it, if it wasn't , there was an infinite sequence that converge to a point of $S$ but S dont have limit points