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How many zeroes does $f(z)=z^4+e^z$ have in the unit disc?

ADDED: can you calculate them?

Here the same question is asked about the disk of radius $2$. It can be solved easily by Rouché's theorem since when $|z|=2$, if $z=x+iy$ then $|e^z|=e^x\leq e^2$ so $$|f(z)-z^4|=|e^z|\leq e^2\leq 9<16=|z^4|$$ therefore there are $4$ roots in that disc (up to multiplicity).

Now when $|z|=1$ we don't have this inequality, nor can we use the same trick by subtracting $e^z$ instead, since when $|z|=1$, $1/e\leq |e^z|\leq e$, one side is less than $1$ and the other greater.

This can be used to show that there are no roots in the right half of the disc, where by going around a curve approximating the boundary of the right half of the disc, so always $x>0$, we get $e^x>1\geq |z|$. A direct calculation also shows no roots are on the $y$ axis.

Any ideas on how to deal with the left half?

Community
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Ur Ya'ar
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  • if $z$ is a root such that $|z|<1$ then $|z^5-z^4-1|\leq 1$. Maybe this helps... – Gabriel Romon Jul 30 '16 at 08:09
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    Maybe use cauchys argument principle. – MrYouMath Jul 30 '16 at 08:12
  • "therefore there are 4 roots in that disc (up to multiplicity).". Correct. Now solve the equation $f(z)=0$ (using $W$) and verify that none of the (infinitely many) roots lie on the unit circle. –  Jul 30 '16 at 08:32
  • you need to show that $2 \pi (n-1) < |\int_{|z| = 1} \frac{4 z^3 + e^z}{z^4+e^z} dz| < 2 \pi (n+1)$ – reuns Jul 30 '16 at 09:29
  • @LeGrandDODOM How do you propose to use this fact? – Ur Ya'ar Jul 30 '16 at 13:46
  • @YiannisGalidakis I don't understand your suggestion. What is $W$? My problem is to solve $f(z)=0$... – Ur Ya'ar Jul 30 '16 at 13:48
  • @user1952009 Well I don't really know how to deal with this integral, it seems even more problematic. – Ur Ya'ar Jul 30 '16 at 13:49
  • @UrBen-Ari-Tishler see my edit to JeanMarie's answer – reuns Jul 30 '16 at 15:00
  • @user1952009 but how did you plot it? using some approximation or by some calculation? As I said, I don't know how to calculate this integral. I understand why this would give me the answer (by the argument principle) but I need a way to estimae it "manually". – Ur Ya'ar Jul 30 '16 at 15:45
  • @UrBen-Ari-Tishler all you need is a bound on $\frac{f'(e^{it})}{f(e^{it})}$, then you can sample $\log f(e^{it})$ and 'unwrap' its imaginary part as I did. Here it is obvious that sampling $t$ every $0.01$ is enough, so it proves there are $2$ zeros (with multiplicity) inside the unit disk – reuns Jul 30 '16 at 15:50
  • @user1952009 what do you mean by "unwrapping"? How do you obtain a bound? As I said, I need a manual proof, so sampling 600 times is no good. – Ur Ya'ar Jul 30 '16 at 16:26
  • @UrBen-Ari-Tishler you need to read a course on the argument principle, and the phase unwrapping means making $arg f(e^{it})$ continuous when drawing it for $t \in [0,2\pi)$ – reuns Jul 30 '16 at 16:31
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    @UrBen-Ari-Tishler "Phase unwrapping" is a classical term in signal processing for "keeping memory of the number of turns one has made around a point", instead of restarting at $0$ every time one has reached $2 \pi$... – Jean Marie Jul 30 '16 at 20:12

1 Answers1

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This is not a solution, just a graphical representation of the modulus $|f(z)|$ (with colors tending to red when values tend to zero). This shows that there are two roots inside the unit disk with conjugate (approximate) values $-0.67003127469869344 \ \pm 0.5161249765067623 i$ (the two others are outside the unit disk). enter image description here

Another way to be convinced graphically of the fact that there are 2 zeros inside the unit circle is by plotting $\frac{\text{arg }f(e^{it})}{2 \pi}$ (unwrapped) (using the argument principle) .

enter image description here

Jean Marie
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  • @ UrBen-Ari-Tishler Thanks for having provided the second graphic which is another way of "seeing" the situation. – Jean Marie Jul 30 '16 at 15:33
  • It wasn't me who added the second plot – Ur Ya'ar Jul 30 '16 at 15:46
  • So, I have lost the trace of the unknown contributor who is welcome to say "it's me". I always appreciate co-operation. – Jean Marie Jul 30 '16 at 15:50
  • @JeanMarie my graphic is a proof, together with a bound on $\frac{f'(e^{it})}{f(e^{it})}$ and the fact I used 600 sample points – reuns Jul 30 '16 at 15:51
  • Is a graphic a proof, I don't think so... It is no more a proof than my own graphics. – Jean Marie Jul 30 '16 at 15:55
  • @JeanMarie it is because you don't know the argument principle so well, all we need is to show the closest integer to the unwrapped $arg$ is $2$ – reuns Jul 30 '16 at 15:57
  • @user1952009 I don't want to polemicate. But one thing is to "see" something and be 100% convinced that this is true, another thing is to prove it by clean analysis tools (majorations etc...). I would like to have the opinion of other people on this issue. – Jean Marie Jul 30 '16 at 16:02
  • @JeanMarie you are boring... I told you what you need to make it rigorous : a bound for $f'/f$ on the unit circle, as I said here it is obvious it works (I didn't compute an explicit bound for $f'/f$ because I know looking at my graphic it will work, but yes you need to compute it and it will be rigorous) – reuns Jul 30 '16 at 16:04
  • @user1952009 All right, I am boring. You have just pointed out in your parenthesis what is missing... – Jean Marie Jul 30 '16 at 16:09
  • @JeanMarie I told it before – reuns Jul 30 '16 at 16:10