2

I am trying to show that $ f(z) = ze^{\lambda - z} - 1$, $\lambda > 1$ has a real root inside the disk. I have already showed, using Rouche's Theorem, that there is exactly one root inside the disk. But I am not sure how to show that this root is real.

I have considered the real-valued analog: $xe^{\lambda - x} - 1$ and tried appealing to calculus. I was able to show that this function is increasing on the interval $(-\infty, 1)$. So If I could find two values in this interval such that $f(x) < 0$ and $f(x) > 0$, I would be done by appealing to the intermediate value theorem.

Any tips?

Note that I have already looked at Show $z e^{\lambda-z}-1$ has only one real root in the unit disk. but that post does not discuss how to show the root is real.

Nicholas Roberts
  • 1,106
  • Maybe try to write out what $f(x+iy)$ is and try to deduce that $y$ has to be $0$ by for example looking at real or imaginairy parts of this equation. – Stan Tendijck Mar 29 '19 at 01:05
  • 3
    Other suggestion, which builds upon your own comment. $0$ and $1$ satisfy the conditions, right? – Stan Tendijck Mar 29 '19 at 01:06
  • 1
    @StanTendijck Yes, I see that $f(0) < 0$ and $f(1) > 0$. So by IVT, there must be a root between $0$ and $1$. Thanks for the tip! – Nicholas Roberts Mar 29 '19 at 01:10
  • My first instinct was to try to argue that $\overline{z}$ would be a root whenever $z$ is, but I'm pessimistic about that being fruitful given the structure of the equation. Why not stick with the calculus approach? If you show that it has a root in $(-1,1)$ via calculus, then you're done. – Gary Moon Mar 29 '19 at 01:12
  • @GaryMoon, Is it legitimate to use the intermediate value theorem on the real-valued function $xe^{\lambda - x} - 1$ to show that it has a root and then this will imply the complex valued function $ze^{\lambda - z} - 1$ will have a real root? – Nicholas Roberts Mar 29 '19 at 01:16
  • 1
    Sure. Say that you use the IVT to find a root $\tilde{x}$ of $xe^{\lambda - x} - 1$ with $\lvert \tilde{x} \rvert < 1$. Then, letting $z=\tilde{x} + i0$ will give you a (real) root in $\mathbb{C}$. Since there is only one root in the unit disk, then the $z$ above must be it. – Gary Moon Mar 29 '19 at 01:49
  • 1
    Since $\lambda$ is real, the Taylor coefficients of $f(z) = ze^{\lambda - z} - 1=z\Sigma{\frac{(\lambda-z)^n}{n!}}-1$ are real, so $f(\bar{z})=\bar{f(z)}$, so if $f(z)$ is real, $f(z)=f(\bar{z})$, so in particular if $f(z)=0, f(\bar{z})=0$, so by unicity you are done – Conrad Mar 29 '19 at 02:53
  • 1
    @Conrad Why not an official answer? By the way, you do not need the Taylor series of $f$, it is well known that $\overline{e^w} = e^\overline{w}$. – Paul Frost Mar 29 '19 at 16:53

0 Answers0