Describe the approximate locations of the zeros of the function $$ f(z) = e^{iz}+e^{-iz}+e^z $$ lying outside the circle $|z|=R >>1$.

Another prelim problem. For Rouche's theorem we need to find a bounded domain. I tried for $R<|z|<r$ and then looked in the left half plane so that $|e^z|<1$ but then I cannot apply the triangle inequality to the other side.



The main idea.

Asymptotically, the zeros will lie on the outer perpendicular bisectors of the sides of the convex hull of the exponent coefficients $i$, $-i$, and $1$. In the image below, the border of the convex hull of these points is shown in blue and the perpendicular bisectors of its sides are shown in red.

enter image description here

Some heuristic reasoning.

For $0 < \theta \leq \pi/2$ we have

$$ |e^z| \leq \left|e^{|z|\, e^{i(\theta+\pi/2)}}\right| = e^{|z|\cos(\theta+\pi/2)} $$

for all $z$ with $|\arg z| \geq \theta + \pi/2$, as can be seen from the following picture.

enter image description here

It follows that

$$ f(z) = 2\cos z + O\left(e^{|z|\cos(\theta+\pi/2)}\right) \tag{1} $$

as $|z| \to \infty$ with $|\arg z| \geq \theta + \pi/2$, so that in this sector $f(z)$ looks like a cosine plus something very small asymptotically. We would then expect the zeros of $f(z)$ to approximate the zeros of the cosine as $|z|$ increases in this region.

Similarly since

$$ f\left(w e^{\pm i\pi/4}\right) = 2 e^{w/\sqrt{2}} \cos\left(w/\sqrt{2}\right) + \exp\left(- e^{\mp i\pi/4} w\right) $$

we can show that, for $0 < \theta \leq \pi/2$,

$$ f\left(w e^{i\pi/4}\right) = 2 e^{w/\sqrt{2}} \cos\left(w/\sqrt{2}\right) + O\left(e^{-|w|\sin \theta}\right) \tag{2} $$

as $|w| \to \infty$ with $-\pi/4 + \theta \leq \arg w \leq 3\pi/4 - \theta$ and

$$ f\left(w e^{-i\pi/4}\right) = 2 e^{w/\sqrt{2}} \cos\left(w/\sqrt{2}\right) + O\left(e^{-|w|\sin \theta}\right) \tag{3} $$

as $|w| \to \infty$ with $-3\pi/4 + \theta \leq \arg w \leq \pi/4 - \theta$.

To summarize, in each of the sectors of width $\pi-2\theta$ symmetric about the perpendicular bisectors (the red lines in the first figure), the quantity $f(z)$ looks like a cosine (perhaps times an exponential) plus something that vanishes exponentially as $|z| \to \infty$. For $\theta$ small these sectors actually overlap, so we would expect that the only large zeros of $f(z)$ are those that approximate the zeros of these cosines.

Between the bisectors.

Let's show that $f(z)$ has no zeros between the perpendicular bisectors for $|z|$ large enough.

First we consider the sector $|\arg z| < \pi/4$, where the dominant term of $f(z)$ is $e^z$. Let $\theta > 0$ be small and fixed. We have

$$ e^{-(1-i)z} = O\left(e^{-\sqrt{2}|z|\sin\theta}\right) \quad \text{and} \quad e^{-(1+i)z} = O\left(e^{-\sqrt{2}|z|\sin\theta}\right) $$

as $|z| \to \infty$ with $|\arg z| \leq \pi/4 - \theta$, from which it follows that

$$ f(z) = e^z \left(1 + e^{-(1-i)z} + e^{-(1+i)z}\right) = e^z \left[1 + O\left(e^{-\sqrt{2}|z|\sin\theta}\right)\right] $$

$|z| \to \infty$ with $|\arg z| \leq \pi/4 - \theta$. We may conclude from this that $f(z) \neq 0$ in the region $|\arg z| \leq \pi/4 - \theta$ for $|z|$ large enough.

Similar calculations yield the analogous result for the remaining two sectors.

Locating the zeros with Rouché's theorem.

The cosine function has zeros at $z = -\pi(n+1/2)$ for $n \in \mathbb Z$. The ones we are interested in lie on the negative real axis (corresponding to $n \geq 0$).

One can show that, for $n \in \mathbb Z$,

$$ |\cos z| > \frac{2}{\pi} \left|z + \pi\left(n + \frac{1}{2}\right)\right| $$

for all $z$ satisfying $\left|z + \pi\left(n + 1/2\right)\right| < \pi/2$. We will estimate the various terms of our function on circles of radius $R_n$ surrounding each of the zeros $-\pi(n+1/2)$, so to use this fact we will assume that $0 < R_n < \pi/2$.

Now, on the circle $\left|z + \pi\left(n + 1/2\right)\right| = R_n$ we have

$$ |e^z| \leq e^{-\pi(n+1/2)+R_n} $$


$$ \frac{4}{\pi} R_n = \frac{4}{\pi} \left|z + \pi\left(n + \frac{1}{2}\right)\right| < |2\cos z|. $$

To connect these we can choose $R_n$ to satisfy

$$ e^{-\pi(n+1/2)+R_n} = \frac{4}{\pi} R_n, $$


$$ R_n = - W\left(-\frac{\pi}{2} e^{-\pi(n+1/2)}\right), $$

where $W$ is the principle branch of the Lambert $W$ function. Since

$$ -W(-z) = z + O(z^2) $$

as $z \to 0$ this implies that

$$ R_n \sim \frac{\pi}{2} e^{-\pi(n+1/2)} $$

as $n\to\infty$. We have met the requirements of Rouché's theorem and may now conclude that

The function $f(z)$ has a zero $z_n$ of the form $$ z_n = -\pi\left(n + \frac{1}{2}\right) + O\left(e^{-\pi n}\right) $$ as $n \to \infty$.

A similar argument will yield

The function $f(z)$ has a zero $z_n$ of the form $$ z_n = \sqrt{2} \pi e^{\pm i \pi/4} \left(n+\frac{1}{2}\right) + O\left(e^{-\sqrt{2} \pi n}\right) $$ as $n \to \infty$.

If you'd like you can go further by choosing the $R_n$ to all be the same size to show that any other zeros, if they exist, must lie "between" these roots. Then show to the contrary that $f(z)$ is eventually nonzero in any ball of fixed radius there which overlaps with the $R_n$ balls. From this you may conclude that the zeros described above are the only large zeros of $f(z)$.

Further reading.

For a general result for zeros of finite sums of the form

$$ \sum c_k e^{\lambda_k z} $$

you may wish to look at Section 2 of the nice paper Zeros of sections of exponential sums by Bleher and Mallison. I followed the basic outline of their results in writing this answer. They also give an older reference which treats a slightly more general case.

  • $\begingroup$ Congrats for all the work, +1. $\endgroup$ – Julien Apr 25 '13 at 23:20
  • $\begingroup$ @julien thanks, it ended up begin much longer than I thought it would when I started typing it! $\endgroup$ – Antonio Vargas Apr 25 '13 at 23:25
  • $\begingroup$ Do you think there could possibly be some shortcuts, given that it seems to be from an exam. I actually don't know what prelim problem refers to, neither of what difficulty this can be... $\endgroup$ – Julien Apr 25 '13 at 23:28
  • $\begingroup$ @julien yes, it's very likely. The application of Rouché's theorem at least can be packed up into a "ready to go" theorem if you're okay with Big-O error estimates in the end (this is what Bleher and Mallison do). I think you could probably jump straight from formulas $(1)$, $(2)$, and $(3)$ to the result. $\endgroup$ – Antonio Vargas Apr 25 '13 at 23:43
  • $\begingroup$ @julien Prelims are kind of like checkpoints in some PhD programs. They're administered by the department to make sure you meet some base competency in various subjects (usually there's at least one for algebra and one for analysis). They're taken sometime in the first two years. This is for Canada and US, not sure about elsewhere. $\endgroup$ – Antonio Vargas Apr 25 '13 at 23:48

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.