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Question: Find the number of zeroes of the polynomial $h(z)=z^{5}+5z^{3}+2z^{2}+4z+1$ in the right half-plane.

Comments: There may be a number of ways to arrive at a solution to this problem, but it would be instructive for me to know if anyone can solve it using Rouché's theorem (or the principle of the argument if that proves impossible). My idea is to count the zeros on an area bounded by the imaginary axis and a half-circle $C$ on the right half-plane with center in $z=0$ and of radius $R$ where $R$ is a large enough number to contain all the zeros in the right-half plane. To do so using Rouché, I would need to see if there is a $g(z)$ which has an absolute value larger than $|f(z)-g(z)|$ on the boundary (that is, on the imaginary axis and on $C$). I am unsure however which $g(z)$ to choose.

This is a problem from an old complex analysis exam, it is similar to this problem which I posted earlier. All input appreciated.

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Sid
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1 Answers1

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Choose $g(z) = z^5 + 1$. The zeros of $g$ are $e^{k\pi i/5},\; k \in \{1,3,5,7,9\}$, two of which lie in the right half plane.

Then a slightly generalised version of Rouché's theorem tells you that $h$ also has two zeros in the right half plane.

The standard formulation of Rouché's theorem demands that $\lvert h-g\rvert < \lvert h\rvert$ (or $\lvert h-g\rvert < \lvert g\rvert$) on the boundary $\partial V$ of the region, but looking at the proof, whose main part is the observation that the number of zeros of $f_\lambda = g + \lambda(h-g)$ in $V$,

$$N_\lambda = \frac{1}{2\pi i}\int_{\partial V} \frac{g'(z) + \lambda\bigl(h'(z) - g'(z)\bigr)}{g(z) + \lambda\bigl(h(z) - g(z)\bigr)}\, dz$$

is independent of $\lambda \in [0,\,1]$, since no $f_\lambda$ has a zero on $\partial V$.

The role of the condition $\lvert h-g\rvert < \lvert g\rvert (\text{ or } \lvert h\rvert)$ on $\partial V$ is solely to guarantee that $g + \lambda(h-g)$ also has no zeros on $\partial V$. If we can determine that in a different way, the conclusion still holds.

For a large enough semicircle in the right half plane, we have $\lvert 5z^3 + 2z^2 + 4z\rvert < \lvert z^5 + 1\rvert$ since the RHS is a polynomial of higher degree than the LHS.

On the imaginary axis, although $\lvert g-h\rvert > \lvert g\rvert$ on some parts, expanding shows

$$g(it) + \lambda\bigl(h(it) - g(it)\bigr) = 1 + it^5 + \lambda(-5it^3 - 2t^2 + 4it) = (1-2\lambda t^2) + it(t^4 - 5\lambda t^2 + 4\lambda),$$

and the real part vanishes only for $\lambda = \dfrac{1}{2t^2}$ (with $t^2 \geqslant \frac12$, since $\lambda \leqslant 1$), when the imaginary part is

$$t\left(t^4 - \frac{5}{2} + \frac{2}{t^2}\right) = \frac1t\left(t^6 - \frac{5}{2}t^2 + 2\right),$$

and the polynomial $x^3 - \frac52x + 2$ has no positive real roots, so we conclude $g(z) + \lambda(h(z)-g(z)) \neq 0$ on the imaginary axis.

Daniel Fischer
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  • Nice solution. I suppose that if you want to use Rouché's theorem then there is no better way. I wonder if a simpler solution exists though which I could use for this type of problems on my exam. I thought of using the principle of the argument, but in order to calculate the integral of $\frac{f'(z)}{f(z)}$ over the curve using residue calculus, then I would have to know the zeros of $f(z)$ which is what I wanted to find to begin with. If you can think of any relatively simple method I would appreciate it, otherwise I thank you for this solution. – Sid Aug 29 '13 at 15:23
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    You can also use intermediate steps to get the classical form of Rouché's theorem. Something like $z^5+5z^3+2z^2+4z+1 \leadsto z^5 + 4z^3 + 2z^2 + 3z + 1 \leadsto \dotsb$ (I didn't check that $\lvert z^3+z\rvert < \lvert h(z)\rvert$ on the imaginary axis, but I'd be surprised if not). Or $g_k = h + \frac{k}{n} (g-h)$; for $n$ large enough, you always have $\frac1n \lvert g-h\rvert < \lvert g_k\rvert$ on the boundary, unless some $g + \lambda(h-g)$ has a zero on the boundary, but that amounts to exactly what I've done. – Daniel Fischer Aug 29 '13 at 15:40
  • what would you do if some of the roots of g(z) lie on the boundary of your region. For example assume $h(z)$ is a 4th degree polynomial and instead we have $g(z) = z^4+1$. now you've got two roots that lie on the boundary of the region(on the imaginary axis). Do those still count towards our tally of roots or does it mean we need to pick a different polynomial, like $g(z) = z^4+z^2+1$ so that the roots are not on the boundary. I also want to add that after researching these types of problems for a few frustrating hours, yours was one of the most clearly explained. Thanks so much! – iYOA Aug 10 '18 at 07:28
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    @iYOA To apply Rouché's theorem one cannot well use a function $g$ having a zero on the boundary in question (aside, $z^4+1$ is real and $\geqslant 1$ on the imaginary axis; $z^4-1$, $z^6+1$ or various other functions would exhibit the problem). One can do that in some special cases where one can determine how the corresponding zeros of $f_{\lambda}$ move if $\lambda$ increases. If $\zeta = \zeta_0$ is a simple zero of $g=f_0$ on the boundary, we have - at least on some $[0,\varepsilon]$ - a continuous dependence of the zero on $\lambda$, i.e. a continuous map $\lambda\mapsto\zeta_{\lambda}$ … – Daniel Fischer Aug 10 '18 at 10:22
  • such that $\zeta_{\lambda}$ is a simple zero of $f_{\lambda}$. If $\zeta_{\lambda}$ is inside the region for $\lambda > 0$, then this zero should be counted, and if its outside it shouldn't. Analogously if $\zeta_{\lambda}$ first stays on the boundary but then moves into the region or outside. If $g$ has a multiple zero on the boundary, that will generically split into several simple zeros, and one has to count how many of them move in, and how many move out. In simple cases it is feasible to determine the behaviour of the boundary zeros, but often that's very complicated. – Daniel Fischer Aug 10 '18 at 10:22
  • Another strategy that can work is to change the region one considers so that $g$ has no zeros on the boundary of the new region. One needs to ascertain that $h$ has no zero lying in one of the two regions but not in the other (or one must be able to find the difference in the number of zeros of $h$ in the new and old regions). Again, this is occasionally easy to do but generically hard. Sometimes one can find a $\tilde{g}$ that works better. If none of that works, a different method of zero counting, e.g. the argument principle, can help. Have you a specific problem you're trying to solve? – Daniel Fischer Aug 10 '18 at 10:23
  • Yeah I was trying to do this with a degree 4 polynomial over here https://math.stackexchange.com/questions/2875666/how-many-zeros-does-the-polynomial-have-in-the-right-half-plane . There was one answer but they didn't use Rouche's theorem or argument principle and I would've liked to see how either of those techniques worked. I think I get how to use rouches principle now from your post(I'd probably pick something like $z^4+2z^2+2$) but I still don't know how to do it with argument principle. – iYOA Aug 11 '18 at 06:10