I have a holomorphic function $f(z)$ which is "near" $\cos z$ in the sense that as $z\to\pm i\infty$, $f(z)$ is dominated by the exponential behavior of $\cos z$. For example $f(z)=\cos z+a$ or $f(z)=\cos z+1/z$. I would like to prove that $f(z)$ has an infinite number of zeros.

Intuitively, I would like to argue as follows: Let $\gamma$ be a straight line path from $z$ to $z+2\pi$, where $\Im(z)$ is large positive. Then $f(\gamma)$ traces out a large circle clockwise around the origin with winding number approximately $-1$. If we decrease $\Im(z)$ this path does something complicated, but as $\Im(z)$ becomes large negative it settles back to an approximate circle around the origin, this time traced counterclockwise, so the winding number is now approximately $1$, and at some point the path had to cross the origin in order to change the winding number.

I realize this proof sketch is problematic, because the path isn't closed, so the origin can "escape" the sweep that way. I think there is a standard technique in complex analysis for this but I'm not sure how to apply it. I am being slightly vague about the function $f$ because I'm not sure what properties I need it to satisfy for this theorem to hold, if the ones I have given are insufficient.

  • $\begingroup$ Is this a part of mathlib?... :-P $\endgroup$ – Kenny Lau Aug 11 '18 at 9:53
  • $\begingroup$ Would Rouche theorem be helpful? $\endgroup$ – Kenny Lau Aug 11 '18 at 9:55
  • $\begingroup$ Yes, Rouché's theorem is what I was after. It's still a bit difficult to apply because I need really strong bounds on the perturbation near the real line, where $\cos z$ is small, but I think it is achievable in my situation. Thanks! You should post that as an answer. $\endgroup$ – Mario Carneiro Aug 11 '18 at 10:22
  • $\begingroup$ I actually don't know how to do it lol I just suggested Rouche because it seems to be helpful. $\endgroup$ – Kenny Lau Aug 11 '18 at 10:25
  • $\begingroup$ So maybe you can post an answer. $\endgroup$ – Kenny Lau Aug 11 '18 at 10:25

For a proof in general you need to state the hypothesis more clearly. Guessing what you mean, it seems to me that $f(z)=e^{iz}$ is "dominated by the exponential behavior of $\cos(z)$ as $z\to\pm i\infty$", but it has no zero.

The two explicit examples you give are easy:


There exists $w$ such that $(w+1/w)/2=-a$. We certainly have $w\ne0$, hence there exist infinitely many $z$ with $e^{iz}=w$, hence $\cos(z)=-a$.

Note that if $|a|\ge1$ then this one cannot be done by Rouche. Because if $\cos(p)=0$ there is no bounded open set $U$ with $p\in U$ such that $|a|<|\cos(z)|$ on the boundary. I mention this because it may be relevant to the general case, if you ever determine exactly what the general case is.


Say $p_1,p_2,\dots$ are the zeroes of $\cos(z)$. Say $\gamma_n$ is a circle with center $p_n$ and radius $1$. There exists $\delta>0$ such that $|\cos(z)|\ge\delta$ on $C_n$. If $n$ is large enough then $|1/z|<\delta$ on $C_n$. Rouche.

  • $\begingroup$ In my situation, it turns out my function is actually closer to $\cos z+\log z$, and I'm only concerned with roots on the right, far from the branch cut. I used $\cos z+a$ as my dominating function, chosen so $\log z-a$ is as small as possible in the places where $\cos z$ drops as low as $1$. Of course $\cos z+a$ also has two roots in a rectangular region with width $2\pi$; I use the strip $[2\pi n,2\pi n+2\pi]$ if $\Re a>0$, otherwise $[2\pi n-\pi ,2\pi n+\pi]$, so that the minimum value of $\cos z$ is at least $1$. $\endgroup$ – Mario Carneiro Aug 12 '18 at 1:26
  • $\begingroup$ By the way, the function I wanted to solve was $\sin^{(n)}\arccos^{(n)}z$, you can read the result at Solution set of $\cos(\cos(\cos(\cos(x)))) = \sin(\sin(\sin(\sin(x))))$. $\endgroup$ – Mario Carneiro Aug 16 '18 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.