2
$\begingroup$

I would like to determine the number of zeros of the polynomial $$p(z) = 3z^4 + z^3 + z^2 + z + 4,$$ in the upper right quadrant of $\mathbb{C}$, using Rouchet's theorem. The zeros should be counted with multiplicity.

So I've set $f(z)$ to be $3z^4 + 4$ and $g(z)$ to be $z^3 + z^2 + z$, and I'm using the curve $\gamma_R$ defined as the closed, positively oriented quarter-circle of radius $R > 2$ around $0$, through the upper right quadrant of $\mathbb{C}$.

Next, I want to show that $|f(z)| > |g(z)|$ for all $z$ with $|z| = R$, and then for all $z$ on the real and imaginary axis. But I don't know how to do that.

Then, I'm thinking it should be enough to figure out the number of zeros of $f(z)$ inside $\gamma_R$, which gives the answer $1$, which is also true according to the solutions manual.

So the problem for me is managing to show that $|f(z)| > |g(z)|$ for all $z \space\epsilon\space \gamma_R$.

$\endgroup$
1
$\begingroup$

Since $|3z^4 + 4| \ge 3R^4 - 4$ and $|z^3 + z^2 + z| \le R^3 + R^2 + R$, certainly $|f(z)| > |g(z)|$ will be true when $R$ is large enough. $R > 2$ happens to work but you don't really need to choose a concrete $R$.

You can show that $|f(z)| > |g(z)|$ on the real and imaginary axis using calculus.

You can also solve this using the argument principle directly: the argument is constantly $0$ on the pos. real axis, it behaves like the argument of $3z^4$ on the circle (so it increases by almost $2\pi$ as the angle goes from $0$ to $\pi/2$); and at the single point where $p(z)$ becomes real on the pos. imaginary axis, i.e. $$\mathrm{Im}[3(iy)^4 + (iy)^3 + (iy)^2 + iy + 4] = 0 \; \Leftrightarrow \; y = 1,$$ the value $p(i)$ is positive. So altogether the argument increases by $2\pi$ and the argument principle guarantees exactly one zero. This method tends to be easier to apply than Rouché if you get the hang of it.

Here is a sketch of what I mean: p(z)

$\endgroup$
3
  • $\begingroup$ I realised I'm still not quite sure how to tell that e.g. $|z^3 + z^2 + z| \leq 3R^4+4$. Sorry for accepting your answer too soon, I thought I had it, but I lost it, so to speak. $\endgroup$
    – frej.mh
    Dec 28 '16 at 2:52
  • 1
    $\begingroup$ @frej.mh Triangle inequality tells you $|z^3 + z^2 + z| \le R^3 + R^2 + R.$ Then $R^3 + R^2 + R \le 3 R^4 - 4$ (certainly true for all large enough $R$, and in fact it is true for all $R \ge 2$) $\endgroup$
    – user399601
    Dec 28 '16 at 2:54
  • $\begingroup$ Great, that makes sense, thank you! $\endgroup$
    – frej.mh
    Dec 28 '16 at 15:08
1
$\begingroup$

Note that for $|z| \le 1$ we have $|g(z)| \le 3 |z| \le 3$. Note that for $|z| > 1$ we have $|g(z)| \le 3 |z|^3 $.

Note that for real $x$, we have $f(x)= f(ix)$.

For $x \in [0,1]$ we have $|g(x)| \le 3 |x| \le 3 < 4 \le p(x)$ and $|g(ix)| \le 3 |ix| \le 3 < 4 \le p(ix)$.

For $x >1$ we have $|g(x)| \le 3 |x|^3 < 3 |x|^4 \le p(x)$ and $|g(ix)| \le 3 |ix|^3 < 3 |ix|^4 \le p(ix)$.

Since there is some $R>0$ such that $3 R^4-4 > R^3+R^2+2$ and $R^3> R^2 +R$, it is clear that $|g(z)| < |f(z)|$ for $|z|=R$, and all of the zeroes of $g$ are contained in $|z|<R$.

Hence $|g(z)| < |f(z)|$ for $z$ on the curve $\gamma_R$, and hence $f$ and $p=f+g$ have the same number of zeros 'inside' $\gamma_R$.

Since the zeros of $f$ are ${1 \over \sqrt{\sqrt{3}}}(\pm 1 \pm i)$ we see that there is exactly one root of $p$ in the upper right quadrant.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.