How many zeros inside the disk?

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In summary, the conversation discusses a problem evaluating the integral of f'(z)/f(z) and how to use the argument principle to solve it. It is suggested to use the fact that e < 3 and to majorize the integrand by 7*z^n on the unit circle. Another approach using Rouche's theorem is also discussed. The conversation ends with the discussion of the tricky issue of the integrand wrapping around the origin and crossing a branch cut.
  • #1
NoDoubts
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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).
 
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  • #3
In general,
[tex]\frac{d}{dz} \ln f(z) = \frac{f'(z)}{f(z)}.[/tex]
 
  • #4
NoDoubts said:
but I have a problem evaluating the integral of
f'(z)/f(z).
Where, exactly, are you having trouble?
 
  • #5
adriank said:
In general,
[tex]\frac{d}{dz} \ln f(z) = \frac{f'(z)}{f(z)}.[/tex]

sure. and how does it help?
 
  • #6
Hurkyl said:
Where, exactly, are you having trouble?

well, f'(z)/f(z) = (1863*z^620 - e^z}) / (3*z^621 - e^z).

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
anyone?
 
  • #8
adriank said:
In general,
[tex]\frac{d}{dz} \ln f(z) = \frac{f'(z)}{f(z)}.[/tex]

I thought this was hint enough. An antiderivative of [tex]f'(z)/f(z)[/tex] is [tex]\ln f(z) = \ln(3z^{621} - e^z)[/tex].

Am I missing something?
 
  • #9
  • #10
Maybe the argument principle isn't what you should be using?
 
  • #11
do you know how else I can solve this problem?
 
  • #12
I cheated a bit and made some plots, but it looks like 3zn - ez has n zeros in the unit disk, and they look like they're nearish to the zeros of zn - 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.
 
  • #14
Looks like that does it. :)

I just started taking a complex analysis course, so unfortunately I'm not very well-informed with regards to that stuff. Glad you solved it, though.
 
  • #15
cheers. now I can move on relieved :)
 
  • #16
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.
 

Related to How many zeros inside the disk?

1. How many zeros are inside the disk?

The number of zeros inside a disk can vary depending on the size and type of the disk. A typical hard disk can hold around 10^15 zeros, while a smaller disk like a CD can hold around 10^12 zeros.

2. What do the zeros inside a disk represent?

The zeros inside a disk are used to represent binary data. Each zero represents a bit, which is the smallest unit of information in a computer.

3. Can a disk contain an infinite number of zeros?

No, a disk cannot contain an infinite number of zeros. The maximum number of zeros a disk can hold is determined by its capacity, which is limited by physical constraints.

4. How are zeros stored on a disk?

Zeros are stored on a disk using magnetic fields. When data is written onto a disk, the magnetic fields are changed to represent the binary code of the data. Zeros are represented by a lack of magnetic field.

5. Is the number of zeros inside a disk important?

Yes, the number of zeros inside a disk is important as it determines the storage capacity of the disk. The more zeros a disk can hold, the more data it can store.

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