
#1
Dec1410, 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 ac PLEASE HELP 



#2
Dec1410, 11:49 PM

P: 1,106

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.




#3
Dec1510, 12:03 AM

P: 37

are my answers correct and how can i do (c)?




#4
Dec1510, 12:27 AM

P: 1,106

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) = (za)^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.




#5
Dec1510, 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? 



#6
Dec1510, 12:43 AM

P: 1,106

Looks good, though you don't really need the g.




#7
Dec1510, 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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