Exponent of convergence of the sequence of zeros of $e^z+z$

Question

Question: How to calculate the exponent of convergence of sequence of zeros of the function $f(z)=e^z+z$?

I know the formula (given below) to calculate the exponent of convergence but for this, I need to know the zeros of the function explicitly, which I don't know. I found in the literature that the exponent of convergence of sequence of zeros of the function $f$ is 1, but I can't understand how it is so.

The exponent of convergence is given by $$ \lambda(f)=\inf \left\{p> 0:\sum_{i=1}^{\infty}\frac{1}{|a_i|^p}<\infty\right\} $$ where $a_i$ are the zeros of $f$.

Can someone help me to understand how shall I proceed?

Answer 1

You do not need the full explicit form (which requires nonelementary functions) to assess the convergence of the series. The asymptotic leading behavior is enough.

Render the equation as $\exp(z)=-z$ and take logarithms to obtain

$z=(2n+1)\pi i+\ln(z)$

where the imaginary part of $\ln(z)$ is taken as $\in ]-\pi,\pi]$ and different values of $n\in\mathbb{Z}$ correspond to the different branches of the logarithm. Then as $|z|$ and $|n|$ approach infinity, the logarithm becomes absolutely small compared with $z$ itself and thus the leading behavior reduces to

$z\sim 2n\pi i, |z|,|n|\to\infty.$

Then the series is easily seen to converge iff $p>1$.

Answer 2

Denote $f(z)=z+e^z$. On the real line $f$ is strictly increasing and takes all values, so, it has the unique real root. Also, $f(\bar z)=\overline{f(z)}$, so, other roots are partitioned onto pairs of complex conjugate non-real numbers. Thus, it suffices to consider only roots in the upper half-plane $\{z:\Im z>0\}$. Let $w=re^{it}$ be such a root, where $0<r<\infty$ and $0<t<\pi$. Then $e^w=-w=e^{it+\log r+i\pi}$, thus $w=it+\log r+i\pi+2\pi i k$ for certain integer $k$ that is equivalent to a system of equations $r\cos t=\log r$; $r\sin t=t+\pi+2k$. Since $0<t<\pi$, we get $r\sin t>0$, thus $k\geqslant 0$. For fixed integer $k\geqslant 0$, we have two equations for $r$ and $t$: \begin{align*}r&=\frac{t+\pi+2\pi k}{\sin t}\tag{1}\\\frac{\log r}r&=\cos t.\tag{2}\end{align*} Equation (1) yields $r>\pi/\sin t>\pi$, thus $\log r>0$ and by (2) we get $0<t<\pi/2$. For fixed $r>\pi$ we may determine $t$ from (2) as $t=\arccos \frac{\log r}r=:T(r)$, then (1) holds iff $$\frac{(T(r)+\pi+2\pi k)^2}{r^2}+\frac{\log^2 r}{r^2}=1\Leftrightarrow T(r)+\pi+2\pi k=\sqrt{r^2-\log^2 r}.$$ For large $r$, the function $\sqrt{r^2-\log^2 r}-T(r)$ increases (its derivative is $1+o(1)$) and goes to infinity, thus it takes the value $\pi+2\pi k$ for large enough $k$ exactly once, and $r$ and $k$ are of the same order.

Therefore, root lengths grow linearly, which immediately yields what you need.

Answer 3

Here's a way to approach this directly with complex analysis methods. By symmetry, it suffices to count zeroes in the upper half plane.

Observe that for any $k\in \mathbb{Z}$, along the line $\Im(z)=(2k+\frac{1}{2})\pi$ we have that $e^z$ is a positive real multiple of $i$. In particular, for any $t\in [0,1]$, the function $f_t(z)=e^z + (1-t)z+ti$ has no zero on that line for $k\geq 0$ (as then both $e^z$ and $(1-t)z+ti$ lie strictly in the upper half plane).

Now recall that $\frac{1}{2\pi i}\int_\gamma \frac{df_t}{f_t(z)}$ counts the number of zeroes of $f_t$ enclosed by the contour $\gamma$. If we let $\gamma$ be the boundary of the rectangle where $(2k+\frac{1}{2})\pi\leq \Im(z)\leq (2(k+1)+\frac{1}{2})\pi$ and $\Re(z) \in [-C_1,C_2]$, then by the above $f_t(z)$ has no zeroes on the horizontal boundary lines of this rectangle. But since $\frac{1}{2} \leq |(1-t)z+ti| \leq \max(|z|,1)$ we find $C_1$ and $C_2$ depending on $k$ ($C_1$ is roughly $\ln(2)$, and $C_2$ is roughly $\ln(2k\pi)$) for which $f_t(z)$ also has no zeroes on the vertical boundary lines of this rectangle. So $\frac{1}{2\pi i}\int_\gamma \frac{df_t}{f_t(z)}$ is an integer depending continuously on $t$, and so agrees for $t=0$ and $t=1$. For $t=0$ we get the number of zeroes in our rectangle for $f_0(z)=e^z+z$, while for $t=1$ we get $f_1(z)=e^z+i$. This has exactly one zero in our rectangle, $(2k+1)\pi i$. Since this argument works for all larger $C_1,C_2$ as well, it follows that $e^z+z$ has a unique zero in each of the horizontal strips $(2k+\frac{1}{2})\pi\leq \Im(z)\leq (2(k+1)+\frac{1}{2})\pi$, with real value bounded logarithmically by $k$. In particular, the absolute value of zeroes grows linearly.