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

Background:

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.

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\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*}

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

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

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  • 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
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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 $1+e^{-1}\le e$.

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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 $$ -i\ln(y_k)+y_k+w=-iz=-ie^{z}=e^{-i\frac\pi2+z}=y_ke^{iw}\\ w=g(w)=-i\,Ln\left(1+\frac{w-i\ln(y_k)}{y_k}\right) $$

Contractivity

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 $k\ge 1$ $$ |g(w)|\le \frac{1+\ln(y_k)}{y_k-1-\ln(y_k)}<0.64<1. $$

Conclusion

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 $(\ln y_k+iy_k)+\Bbb D$.

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