Prove that $\log|e^z-z|\leq |z|+1$ where $z\in\mathbb{C}$ with $|z|\geq e$.


This is from a proof that $e^z-z$ has infinitely many zeroes. The present stage is that we assumed in contradiction that $e^z-z$ hasn't any zero.

My attampt:

I assume that the meaning of $\log$ here is the principal branch of $\log$.

We know that $|w|\in\mathbb{R} ,\ \forall w\in\mathbb{C}$. Because $\log$ is increasing in $\mathbb{R}^+$ and according to the triangle inequality we get $$\log|e^z-z|\leq\log(|e^z|+|z|)$$ But I'm not sure how to proceed. Thanks.


\begin{align*} |e^z-z|&\le|e^z|+|z|\\ &\le e^{|z|}+|z|\quad\textrm{from series expansion}\\ &\le e^{|z|+1}\quad\textrm{again from series expansion} \end{align*}


Use the definition: $e^z = \sum_{n=0}^\infty z^n/n!$

$$ \begin{align} \vert e^z - z \vert &= \vert 1+z^2/2 + z^3/6 + \cdots \vert \\ &\leq 1+\vert z \vert ^2/2 + \vert z^3 \vert/6 + \cdots\\ &= e^{\vert z \vert} - \vert z \vert \\ &\leq e^{\vert z \vert}\\ &< e^{\vert z \vert + 1} \end{align} $$

  • 2
    $\begingroup$ The right hand in the first inequality is wrong: it must be $+$ there instead of $-$ . $\endgroup$ – DonAntonio Jun 22 at 10:34
  • $\begingroup$ @DonAntonio It is in fact true. I reverted to my original answer since people didn't seem to buy it $\endgroup$ – Philip Hoskins Jun 22 at 10:38
  • $\begingroup$ You still would have to justify the very last inequality. $\endgroup$ – DonAntonio Jun 22 at 10:50
  • $\begingroup$ Maybe if OP really wants me to, but I suspect they can figure that one out on their own if they care to. $\endgroup$ – Philip Hoskins Jun 22 at 10:55
  • $\begingroup$ It would be perhaps clearer to replace $e^{|z|}-|z|$ with merely $e^{|z|}$. $\endgroup$ – A.Γ. Jun 24 at 10:40

Hint: estimate for $r=|z|\ge 0$ $$ \log|e^z-z|\le\log(e^r+r)=\log(e^r[1+re^{-r}])=r+\log(1+re^{-r}). $$ Prove that the function $f(r)=re^{-r}$ attains maximum at $r=1$ and $1+e^{-1}\le e$.


There exist a direct quick solution to the problem behind the question, and a solution using the logarithmic instead of the exponential approach that is probably also easier to handle in using the Rouché theorem.

Quick solution using the poly-log or Lambert-W functions

Roots of $e^z-z=0$ come in conjugate complex pairs. As the equation is equivalent to $-ze^{-z}=-1$, these roots can be written with the Lambert-W function as $z_k=-W_{k}(-1)$.

for k in range(-5,5): print k, -lambertw(-1,k)
-5 (3.287768611544094 +26.580471499359145j)
-4 (3.020239708164501 +20.272457641615222j)
-3 (2.6531919740386973+13.949208334533214j)
-2 (2.062277729598284 + 7.588631178472513j)
-1 (0.3181315052047642+ 1.3372357014306893j)
 0 (0.3181315052047642- 1.3372357014306893j)
 1 (2.062277729598284 - 7.588631178472513j)
 2 (2.6531919740386973-13.949208334533214j)
 3 (3.020239708164501 -20.272457641615222j)
 4 (3.287768611544094 -26.580471499359145j)

Long solution using a fixed-point argument

Now one could ask for the existence of these infinitely many branches of the Lambert-W function and a more precise location of the roots.

Approximate root location

For any root $z=x+iy$ satisfying $z=e^z$ we get $$x^2+y^2=e^{2x}.$$ First this means that $e^x>|x|$ or $x>-W_0(1)=-0.56714329...$. For large $x$ we have $e^x\gg x$ so that the value of $y$ has to dominate, $y\sim\pm e^x$. Then the phase of $z$, as $z$ is almost on the imaginary axis, is $\pm\frac\pi2$. As that has to be reflected in the imaginary part of the exponent in $e^z$, this requires that for the root in the upper half-plane $Im(z)\approx y_k= 2\pi k+\frac\pi2$, $k>0$.

Then a first approximation for the root is $$z\approx \ln(y_k)+iy_k=\ln(2\pi k+\frac\pi2)+i(2\pi k+\frac\pi2).$$

Construction of a fixed-point function

It appears thus advisable to orient the corrections to this approximation in the same direction, that is, set $z=\ln(y_k)+iy_k+iw$, so that $$ -i\ln(y_k)+y_k+w=-iz=-ie^{z}=e^{-i\frac\pi2+z}=y_ke^{iw}\\ w=g(w)=-i\,Ln\left(1+\frac{w-i\ln(y_k)}{y_k}\right) $$


The fixed-point map $g$ has, for $k\ge 1$, the derivative bound $$ g'(w)=\frac{-i}{y_k-i(w+\ln(y_k))}\implies |g'(w)|<\frac1{2\pi k} ~~\text{for }|w|\le1 $$

Self-mapping property

Further, from $|Ln(1+v)|\le\frac{|v|}{1-|v|}$ is obtained that for $|w|\le1$ and $k\ge 1$ $$ |g(w)|\le \frac{1+\ln(y_k)}{y_k-1-\ln(y_k)}<0.64<1. $$


This means that the unit disk $\Bbb D$ is mapped into itself and the mapping $g$ is contractive on the unit disk, so that there is exactly one fixed-point, meaning exactly one root of the equation in the translated disk $(\ln y_k+iy_k)+\Bbb D$.


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.