Let $\lambda > 1$ , want to show that the equation $$\lambda-z-e^{-z}=0$$ has exactly one solution in the right half plane $\{z:Re(z)>0\}$. Moreover, the solution must be real.

I tried to use Rouche's theorem on $g(z)=\lambda - z$ and $f(z)=e^{-z}$ to get that the number of zeros of $f+g$ and the number of zeros of $g(z)$ is the same, and since $g(z)$ has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve $\gamma$ such that this will work.

for the second part I used the IVT to show that $\lambda -x-e^{-x}$ has a zero in $(0,\lambda)$ to conclude that the solution is real. is this acceptable? Thank you for your help.

  • $\begingroup$ How did you use Roche's theorem? Don't you need $|f(z)| < |g(z)|$ for say $z = \lambda$. $\endgroup$
    – muzzlator
    Mar 4 '13 at 2:15
  • $\begingroup$ @muzzlator I took a small circle with center $\lambda$ and the inequality you wrote should be true for all values on (not in) this circle, and this is that case. $\endgroup$
    – i.a.m
    Mar 4 '13 at 2:19
  • $\begingroup$ Ah, I see. Yeah that looks like it works $\endgroup$
    – muzzlator
    Mar 4 '13 at 2:23
  • $\begingroup$ Rouche is a good idea. But if you apply it to a small circle, how does this apply to the whole half-plane? $\endgroup$
    – Julien
    Mar 4 '13 at 2:56
  • $\begingroup$ @julien I was thinking using part two i.e the real case we established that the equation has a zero in the right half plane, now lets assume we have another zero in the right half plane and take a circle containing both zeros to get a contradictoin, and conclude that there is exactly one solution. the problem I'm facing now is that I can't get the inequality of Rouche's theorem. $\endgroup$
    – i.a.m
    Mar 4 '13 at 3:02

A hint.

If $\operatorname{Re} z > 0$ and $\lambda - z - e^{-z} = 0$ then

$$ |\lambda - z| = e^{-\operatorname{Re} z} < 1. $$

In other words, if the equation has any solutions in the right half-plane then they lie in the open disc $|z-\lambda|<1$.

  • 3
    $\begingroup$ so if I take $\gamma$ to be the circle $|\lambda-z|=1$ then we have $|f(z)|=e^{-Re(z)}<1=|\lambda-z|=|g(z)|$, Thus, by Rouche's theorem $g$ and $f+g$ have the same number of zeros in this circle which is 1. $\endgroup$
    – i.a.m
    Mar 4 '13 at 15:07
  • 2
    $\begingroup$ @i.a.m You're welcome. $\endgroup$ Mar 4 '13 at 15:57
  • $\begingroup$ wait, is the question asking us to prove there is only one solution which happens to be in right half plane OR that there exists only 1 solution in right half plane (and possible some solution in left half plane)? $\endgroup$ Sep 9 '18 at 7:54
  • $\begingroup$ @AlphaRomeo, the equation has infinitely-many solutions, and exactly one of those is in the right half-plane. $\endgroup$ Sep 9 '18 at 20:48
  • $\begingroup$ Can you please explain why you say infinitely-many solutions in left half plane? If z is purely real then I plotted and checked it can have only 1 solution. How to analyze when z is not purely real? $\endgroup$ Sep 10 '18 at 16:02

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