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How many zeros inside the disk? 
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#1
Feb309, 04:24 AM

P: 20

How many zeros inside the unit disk does the following function have?
f(z) = 3 z^621  e^z. (in words, three times z to the power 621 minus e^z :) ...argument principle answers this question, but I have a problem evaluating the integral of f'(z)/f(z). 


#2
Feb309, 08:24 PM

P: 20



#3
Feb309, 11:54 PM

P: 534

In general,
[tex]\frac{d}{dz} \ln f(z) = \frac{f'(z)}{f(z)}.[/tex] 


#4
Feb409, 12:17 AM

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PF Gold
P: 16,099

How many zeros inside the disk?



#5
Feb409, 01:12 AM

P: 20




#6
Feb409, 01:17 AM

P: 20

Now, we need to integrate that along the contour C. I've tried to integrate it directly by setting z = e^{theta*i) where 0 < theta <= 2pi , but got stuck. 


#7
Feb409, 09:27 PM

P: 20

anyone?



#8
Feb409, 09:29 PM

P: 534

Am I missing something? 


#9
Feb409, 09:42 PM

P: 20

yes. this is obvious, but how does it answer the question?
what we need to do is, according to argument principle, to calculate an integral on closed countour which is unit circle. http://en.wikipedia.org/wiki/Argument_principle 


#10
Feb409, 10:34 PM

P: 534

Maybe the argument principle isn't what you should be using?



#11
Feb409, 11:35 PM

P: 20

do you know how else I can solve this problem?



#12
Feb509, 12:00 AM

P: 534

I cheated a bit and made some plots, but it looks like 3z^{n}  e^{z} has n zeros in the unit disk, and they look like they're nearish to the zeros of z^{n}  1. It looks like you have to use the fact that e < 3, since replacing the constant 3 with any number greater than e gives the same number of zeros. If you replace it with e, then you get a zero at z = 1.



#13
Feb509, 12:09 AM

P: 20

yep. I think I nailed it by Rouche's theorem http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem
you can majorize the 3zn  ez  by the 7*z^n on the unit circle z=1, then the result follows. 


#14
Feb509, 12:11 AM

P: 534

Looks like that does it. :)
I just started taking a complex analysis course, so unfortunately I'm not very wellinformed with regards to that stuff. Glad you solved it, though. 


#15
Feb509, 12:12 AM

P: 20

cheers. now I can move on relieved :)



#16
Feb509, 12:18 AM

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PF Gold
P: 16,099

For the record, adriank's earlier hint would have let you directly compute the contour integral... except for the sticky problem that the value of the integrand wraps around the origin (621 times!), and thus must cross a branch cut (621 times), so the straightforward thing wouldn't work.
Of course, if you could prove that it wraps around the origin exactly 621 times, and thus crosses the branch cut 621 times, you could take advantage of the fact you know exactly how big the discontinuity is, and so you still could make a direct calculation of the contour integral. 


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