# How many zeros does the function $f(z)=z^8+e^{-2016\pi z}$ have in the region $\mathrm{Re}(z)>0$?

For the function $$f(z)=z^8+e^{-2016\pi z}$$ defined on $\mathrm{Re}(z)>0$, I am tring to find all roots.

I have initally tried finding the roots of the function via Rouché's Theorem. I at first thought this function had no zeros in this region, however, after playing around in Mathematica, I now believe there should be $2024$ distinct roots. I can't think of a function which would satisfy the conditions of Rouché's Theorem and have that many roots, so I don't believe this is probably the best way to go about it.

From this point, I thought perhaps we could use the Argument principle, by it does does not appear the integral will be easy to evaluate. It can be easily shown that all the roots with $\mathrm{Re}(z)>0$ lie within the open half unit disk, so the path around this disk might be best if I went through such an approach.

• If $z$ is such a zero, then $$0= |z^8+e^{- 2016 \pi z}| \ge |z^8| - |e^{- 2016 \pi z}| \ge |z|^8 - e^{- 2016 \pi |z|}$$ what can you conclude about $|z|$? – Crostul Apr 27 '17 at 23:05
• From here we have $|z|\leq e^{-252\pi Re(z)}$, but I don't see how this will help much, sorry. – Apen13 Apr 27 '17 at 23:22
• You can get from this that all the roots must be in the unit circle. I agree with this. I understand this, but I'm not sure how to use this fact come into the proof, since it seems I need to either find a function which satisfies Rouche's theorem on the boundary of the unit circle and has 2024 roots inside, or calculate the integral of $\frac{f'(z)}{f(z)}$ over the bounds to use the disk to use the argument principle. – Apen13 Apr 28 '17 at 1:58
• @Crostul: You assume that $|z|\le Re(z)$ to get to your estimate. – LutzL Apr 28 '17 at 8:35