Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to show that the equation $az^{3}-z+b=e^{-z}(z+2)$, where $a>0$ and $b>2$, has two roots in the right half-plane $\mathrm{Re}\;z\geq 0$. This is a problem in using Rouche's theorem but I am unable to get the right estimates.

I tried taking a semicircular contour in the right half-plane with its arc going from $-iR$ to $iR$ and then going down the imaginary axis. To apply Rouche's theorem, I need holomorphic functions $f,g$ such that $|g(z)|<|f(z)|$ for $z$ on the semicircle. I took $f(z)=az^{3}-z+b$ and $g(z)=e^{-z}(z+2)$. Things seem to be fine on the arc but I ran into problems on the imaginary axis.

share|cite|improve this question
So... what did you try? – akkkk Dec 25 '12 at 20:12
up vote 2 down vote accepted

I dont think there is a problem on the imaginary axis. You want to find a bound for $$|e^{-z} (z+2)|$$

But $|e^{-z}|=e^{-x}=1$ since $x=0$ on the imaginary axis. Letting $z=iy$ we have $$|g(z)|^2=y^2+4$$ and $$|f(z)|^2=y^2(a y^2+1)^2+b^2$$

It is clear now that $|g(z)|<|f(z)|$.

share|cite|improve this answer
But don't we need the inequality for all points on the contour, in particular all points on the imaginary axis? I was getting problems when considering points on the imaginary axis with small modulus. – user44532 Dec 26 '12 at 14:49
But if you assume that $R$ is large then my inequalities hold for all points on the imaginary axis. Even at $z=0$ you have $|g(0)|=2$ and $|f(0)|=b>2$. – PAD Dec 26 '12 at 18:46
For points on the imaginary axis with small modulus, I was thinking that $|f(z)|\geq b-a|z|^{3}-|z|$ while $|g(z)|\leq|z|+2$ so the desired inequality holds if $|z|$ is small enough that $a|z|^{3}+2|z|<b-2$. I'm just a little concerned with points with modulus perhaps just slightly bigger than that. Maybe I'm being paranoid. – user44532 Dec 27 '12 at 16:17
o.k. I think you have a point. I will make the computation more precise. – PAD Dec 27 '12 at 21:52
I realized I haven't gotten to the answer to the question. After getting the estimate, we know that the number of roots of the equation in the right half-plane is the same as the number of roots of $az^{3}-z+b$ in the right half-plane so how do we see that there are two? I would think it is sufficient to show that the roots there are not real. – user44532 Dec 28 '12 at 16:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.