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?

  • $\begingroup$ if $z$ is a root such that $|z|<1$ then $|z^5-z^4-1|\leq 1$. Maybe this helps... $\endgroup$ Jul 30, 2016 at 8:09
  • 1
    $\begingroup$ Maybe use cauchys argument principle. $\endgroup$
    – MrYouMath
    Jul 30, 2016 at 8:12
  • $\begingroup$ "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. $\endgroup$
    – user127032
    Jul 30, 2016 at 8:32
  • $\begingroup$ you need to show that $2 \pi (n-1) < |\int_{|z| = 1} \frac{4 z^3 + e^z}{z^4+e^z} dz| < 2 \pi (n+1)$ $\endgroup$
    – reuns
    Jul 30, 2016 at 9:29
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    $\begingroup$ @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 $2 \pi$... $\endgroup$
    – Jean Marie
    Jul 30, 2016 at 20:12

1 Answer 1


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

  • $\begingroup$ @ UrBen-Ari-Tishler Thanks for having provided the second graphic which is another way of "seeing" the situation. $\endgroup$
    – Jean Marie
    Jul 30, 2016 at 15:33
  • $\begingroup$ It wasn't me who added the second plot $\endgroup$
    – Ur Ya'ar
    Jul 30, 2016 at 15:46
  • $\begingroup$ So, I have lost the trace of the unknown contributor who is welcome to say "it's me". I always appreciate co-operation. $\endgroup$
    – Jean Marie
    Jul 30, 2016 at 15:50
  • $\begingroup$ @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 $\endgroup$
    – reuns
    Jul 30, 2016 at 15:51
  • $\begingroup$ Is a graphic a proof, I don't think so... It is no more a proof than my own graphics. $\endgroup$
    – Jean Marie
    Jul 30, 2016 at 15:55

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