5
$\begingroup$

I've been practicing some problems on Rouche's theorem, but this one has given me trouble.

How can I find how many zeroes of $z^4+4z^3+6z^2-4z+3$ inside $|z-1|<1$?

The given polynomial looks similar to $(z-1)^4=z^4-4z^3+6z^2-4z+1$, so I let $g(z)$ be the given polynomial, and take $f(z)=8z^3+2$. Then on the circle $|z-1|=1$, $ |f(z)-g(z)|=|(z-1)^4|=|z-1|^4=1. $ If I could show $|f(z)|>1$ on the curve, then I would have that $f$ and $g$ have the same number of zeroes inside the circle, which I think is zero for $f$. Is this the right idea, or have I gone astray? Thanks.

  • 0
    Relevant: [this](http://mathworld.wolfram.com/Schur-CohnAlgorithm.html), [this](http://en.wikipedia.org/wiki/Jury_stability_criterion) and [this](http://en.wikipedia.org/wiki/Rouch%C3%A9's_theorem)2012-03-28

1 Answers 1

3

Two of the four complex roots are inside the unit circle centered at $1$. If you find these two pairs of conjugate roots, you can work out that $ g(z)\approx(x^2+4.64751368404x+8.66302952977)(x^2-0.647513684039x+0.346299177406) $ and that the roots are approximately $ \{ -2.32375684201972 \pm 1.80642842895505 i,~ 0.32375684201972 \pm 0.491406842291623 i \}, $ the former pair of which lie outside and the latter pair inside the given circle. However, I needed a numerical method to obtain the above and haven't used Rouche's theorem; I like your approach...I'll consider how to proceed...

In fact, $f(z)=2(4z^3+1)$ has roots at $-\frac1{\sqrt[3]{4}}\,e^{\frac{2\pi ik}{3}}$ for $k=0,\pm1$ along the circle of radius $2^{-\frac23}$, two of which are inside $|z-1|=1$ since $\left|1+\frac1{\sqrt[3]{4}}\,e^{\pm\frac{2\pi i}{3}}\right|<1.$ (For $k=\pm1$, the roots are on the side of a unit equilateral triangle stemming radially from the origin, which is thus inside the unit circle centered at $1$.)

Also, on the unit circle in question, if we let $z=1+e^{it}$ following @RobertIsrael's lead, then $ \eqalign{ h(t)&= \left| f(z) \right|^2 =4\cdot \left| 4\left(1+e^{it}\right)^3+1 \right|^2 =4\cdot \left| 5+12e^{it}+12e^{2it}+4e^{3it} \right|^2 \\& =4\cdot \left[ \left(5+12\cos{t}+12\cos{2t}+4\cos{3t}\right)^2+4\, \left( 3\sin{t}+ 3\sin{2t}+ \sin{3t}\right)^2 \right] \\& =4\cdot \Big[ (5^2+12^2+12^2+4^2) + 2\cdot5\cdot4\, \left(3\cos{t}+3\cos{2t}+\cos{3t}\right) \Big.\\&\qquad\Big. +2\cdot4^2\cdot3\,\left(3\cos{(2t-t)}+\cos{(3t-t)}+\cos{(3t-2t)}\right) \Big] \qquad\text{(combining pairwise terms)} \\& =4\cdot \left[ 329 + 8\, \left(15\cos{t}+15\cos{2t}+5\cos{3t}\right) + 8\, \left(48\cos{t}+12\cos{2t}\right) \right] \\& =4\cdot \left[ 329 + 8\, \left(63\cos{t}+27\cos{2t}+5\cos{3t}\right) \right] } $ by expanding the squares, combining pairwise terms and using $\cos(a-b)=$ $\cos a\cos b+$\sin a\sin b$ with $a,b\in\{t,2t,3t\}$. Differentiating and using the double & triple angle formulas (or $\href{http://en.wikipedia.org/wiki/Chebyshev_polynomials#Examples}{U_1,~U_2}\in\mathbb{Z}[\cos t]$), $ \eqalign{ h'(t)& =-96\, \left[ 21\sin{t}+18\sin{2t}+5\sin{3t} \right] \\&= -96\,\sin t\, \left[ 21 + 18 \left( 2\cos t \right) + 5 \left( 4 \cos^2t - 1 \right) \right] \\&= -96\,\sin t\, \left[ 20 \cos^2t + 36 \cos t + 16 \right] \\&= -384\,\sin t\, \left( \cos t + 1 \right) \left( 5 \cos t + 4 \right) } $$ we obtain the critical points and find the global minimum as he did in his comment, showing that $|f(z)|\ge\frac65$ on the circle in question.

enter image description here

So now you can use Rouché's theorem.

  • 1
    Not much nudging. You can do something like this: > f := z -> 8*z^3+2: > Q := evalc(abs(f(1+exp(I*t)))^2): > solve(diff(Q,t)=0,t,AllSolutions); Then you recognize that it's enough to take _Z1 to be $0$ or $1$ and the other _Zi parameters to be $0$.2012-03-28