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Complex Analysis Singularities and Poles

by bballife1508
Tags: complex analysis, pole, singularity
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bballife1508
#1
Dec14-10, 11:20 PM
P: 37
Assume throughout that f is analytic, with a zero of order 42 at z=0.

(a)What kind of zero does f' have at z=0? Why?

(b)What kind of singularity does 1/f have at z=0? Why?

(c)What kind of singularity does f'/f have at z=0? Why?



for (a) I'm pretty sure it is a zero of order 41

for (b) I'm almost sure it is a pole of order 42

but for (c) I am not quite sure nor can I really explain any of a-c

PLEASE HELP
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snipez90
#2
Dec14-10, 11:49 PM
P: 1,104
Since you arrived at a plausible answer, you must have some intuition for this. Explain it, and go back to the definitions if you need to. You must have some idea about the form of a holomorphic function with a zero of order 42.
bballife1508
#3
Dec15-10, 12:03 AM
P: 37
are my answers correct and how can i do (c)?

snipez90
#4
Dec15-10, 12:27 AM
P: 1,104
Complex Analysis Singularities and Poles

Okay, if some analytic function f has a zero of order n at z = a, then you can certainly write it as f(z) = (z-a)^n * h(z) where h is holomorphic, right? This much should at least be true even if all you had was a sensible notion of a zero of a function. So go from here to explain a)-c). It is NOT hard to work out the details.
bballife1508
#5
Dec15-10, 12:35 AM
P: 37
is it sufficient to say that

f(z)=z^42*h(z) which implies that f'(z)=42z^41*g(z)

so f'/f has an extra z on the bottom so there is a pole of order 1?
snipez90
#6
Dec15-10, 12:43 AM
P: 1,104
Looks good, though you don't really need the g.
bballife1508
#7
Dec15-10, 12:56 AM
P: 37
what do you mean i don't need the g? i can't use h still correct?


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