# Prove $\log|e^z-z|\leq |z|+1$

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.

\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}

• The right hand in the first inequality is wrong: it must be $+$ there instead of $-$ . – DonAntonio Jun 22 at 10:34
• @DonAntonio It is in fact true. I reverted to my original answer since people didn't seem to buy it – Philip Hoskins Jun 22 at 10:38
• You still would have to justify the very last inequality. – DonAntonio Jun 22 at 10:50
• Maybe if OP really wants me to, but I suspect they can figure that one out on their own if they care to. – Philip Hoskins Jun 22 at 10:55
• It would be perhaps clearer to replace $e^{|z|}-|z|$ with merely $e^{|z|}$. – 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 $$-i\ln(y_k)+y_k+w=-iz=-ie^{z}=e^{-i\frac\pi2+z}=y_ke^{iw}\\ w=g(w)=-i\,Ln\left(1+\frac{w-i\ln(y_k)}{y_k}\right)$$

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