# Why $P(z)\text{sin}(z)+Q(z)\text{cos}(z)$ has only finitely many non-real zeroes?

Consider entire functions (defined and holomorphic on the whole complex plane) of the form: $$f(z)=P(z)\text{sin}(z)+Q(z)\text{cos}(z),$$ where $$P(z)$$ and $$Q(z)$$ are polynomials with real coefficients. They seem to always have only finitely many zeros that don't lie on real axis. Namely, there is always some real $$M>0$$ so that if $$|z|>M$$ and $$f(z)=0$$, then $$z\in\mathbb{R}$$. How can I prove this?

I tried using Cauchy's integral theorem and integral formula, I tried inverting the function or using logarithms, but none worked. Intuitively, I think the point is that in order to split ("split" because the zeroes would obviously be symmetrical around the real axis) infinitely many zeroes that $$\text{sin}$$ and $$\text{cos}$$ have, you (in some way) have to introduce infinitely many saddles (zeros of the derivative), but polynomials can have only finitely many. Does this argument make sense and how could I turn it into a rigorous proof?

• @WillJagy Yes, that's the other question on this forum on which I did find a proof: math.stackexchange.com/questions/4038673/… Commented Feb 25, 2021 at 18:08
• @WillJagy I don't think so. Why would it be? Commented Feb 25, 2021 at 18:21
• @WillJagy Yes. And all zeros of $\text{sin}(z)+\text{cos}(z)$ are real. Commented Feb 25, 2021 at 18:26
• This looks like an interesting question. You can write the equation as $\tan(z) = R(z)$ for some rational function $R$ with real coefficients. The equation $\tan(z) = z$ has only real zeros, see for example math.stackexchange.com/q/388673/42969. I doubt however that your idea for a proof works. The mean-value theorem for real functions (“between two zeros is always a zero of the derivate”) does not hold in the complex plane. Commented Feb 25, 2021 at 19:00
• @MartinR Yes, you have the point. Anyway, this particular situation offers some more confidence to me to say that because of the symmetry that my function has: $f(\overline{z})=\overline{f(z)}$. Commented Feb 25, 2021 at 19:31

I think that your conjecture is correct, and can be proved using Rouché's theorem.

Without loss of generality one can assume that $$\deg Q < \deg P$$, otherwise consider the function \begin{align} f(z +a) &= \left( P(z) \cos(a) - Q(z) \sin(a) \right) \sin(z) + \left( P(z) \sin(a) + Q(z) \cos(a) \right) \cos(z) \\ &= \tilde P(z) \sin(z) + \tilde Q(z) \sin(z) \end{align} with a suitably chosen $$a \in \Bbb R$$.

Then $$f(z) = 0$$ can be written as $$\tag{*} \tan(z) = r(z)$$ where $$r$$ is a rational function with real coefficients and $$\lim_{z \to \infty} r(z) = 0$$. In particular we can find an $$M > 0$$ such that $$\tag{1} |r(z)| < 1/2 \text{ for } |z| \ge M \, .$$

We also need some estimates for the absolute value of the tangent. Using $$\tan(x+iy) = \frac{\sin(2x) + i\sinh(2y)}{\cos(2x) + \cosh(2y)}$$ from https://proofwiki.org/wiki/Tangent_of_Complex_Number we have $$\tag{2} |\tan(x+iy)| \ge \frac{|\sinh(2y)|}{\cosh(2y)+1} \text{ for } y \ne 0 \, ,$$ and $$\tag{3} |\tan(x+iy)| = \frac{|\sinh(2y)|}{\cosh(2y)-1} > 1 \text{ for x = \pi/2 + k \pi with k \in \Bbb Z.}$$

Using $$(2)$$ and increasing $$M$$ if necessary, we have $$\tag{4} |\tan(x+iy)| > \frac 12 \text{for |y| \ge M.}$$

It follows from $$(1)$$ and $$(4)$$ that $$(*)$$ has no solutions with $$\operatorname{Im}(z) \ge M$$.

Now consider the rectangles $$R_k = \{ x+iy \mid \frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi, |y| \le M \}$$ with integers $$k$$ satisfying $$\frac 12 \pi + k\pi > M$$ or $$\frac 32 \pi + k\pi < -M$$.

Then $$(1)$$, $$(3)$$ and $$(4)$$ imply that $$|\tan(z)| > 1/2 > r(z)$$ for all $$z$$ on the boundary of $$R_k$$, and it follows from Rouchés theorem that $$\tan(z) -r(z)$$ and $$\tan(z)$$ have the same number of zeros in $$R_k$$, which is one.

On the other hand it follows from the intermediate value theorem that $$\tan(x) - r(x)$$ has one zero in the interval $$\frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi$$. (This is where we use that the rational function is real on the real axis!)

So we have shown the following:

• $$\tan(z) -r(z)$$ has no zeros with $$\operatorname{Im}(z) \ge M$$.
• For sufficiently large $$|k|$$, $$\tan(z) -r(z)$$ has exactly one zero in each strip $$\frac 12 \pi + k\pi \le x \le \frac 32 \pi + k\pi$$, and that zero is real.

In particular, the equation $$(*)$$ has only finitely many non-real solutions.

• I cheated a bit: $\tan$ has poles on the boundary of $R_k$. But that can be fixed by removing small half-disk around the poles from $R_k$. Commented Feb 25, 2021 at 21:17
• I see your idea. It is nice! I didn't know about Rouche's theorem at all. Do you think you can make some short-cuts using the argument principle? I am just curious to know... Commented Feb 25, 2021 at 21:27
• @donaastor: Rouché's theorem is based on the argument principle: From $|f-g| < |g|$ one concludes that $\int f'/f = \int g'/g$ so that $f$ and $g$ have the same number of zeros inside the contour. It may be possible to use the argument principle directly, but I don't know if that makes the proof simpler. Commented Feb 25, 2021 at 21:39
• Oh, then it is explained well enough. That, about the Rouche's theorem. I just thought that using argument principle directly could avoid playing with poles of $\text{tan}$ because it works fine with poles. And, the number of poles of $f(z)$ is much easier to calculate than the number of zeros. Whatever, the rest I can think myself. Thank you for your ideas! Commented Feb 25, 2021 at 22:01