# Determine the number of zeros of the polynomial

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 $z \space\epsilon\space \gamma_R$.

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:

• 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. Dec 28 '16 at 2:52
• @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 $R \ge 2$) Dec 28 '16 at 2:54
• Great, that makes sense, thank you! Dec 28 '16 at 15:08

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.