# Location of roots of degree six complex polynomial

I want to prove that $z^6 +192 z + 640 =0$ has one root in the first and fourth quadrants and two roots in the second and third quadrants. How can I do this? I have tried some ideas like using Vieta's formulas, but still no result.

• Do you mean one root in each of first and fourth, and two roots in each of second and third? And I suppose you also mean each quadrant to not include the bounding parts of x,y axes? – coffeemath Dec 18 '15 at 21:26
• @coffeemath , Yes , Yes – herashefat Dec 18 '15 at 21:27

## 2 Answers

$f(z) = z^6 + 192 z + 640$ has no roots on the real or imaginary axis. The "Routh–Hurwitz theorem" states that $$p - q = \frac 1\pi \Delta \arg f(iy)$$ where

• $p$ is the number of roots of $f$ in the left half-plane,
• $q$ is the number of roots of $f$ in the right half-plane,
• $\Delta \arg f(iy)$ is the variation of the argument of $f(iy)$ when $y$ runs from $-\infty$ to $+\infty$.

Here $f(iy) = (640 - y^6) + (192 y) i$ and the graph looks like this:

The total variation of the argument is $2 \pi$ and therefore $p - q = 2$.

On the other hand, the total number of roots is $6$, therefore $f$ has $4$ roots in the left half-plane and $2$ roots in the right half-plane.

Finally, $f$ has real coefficients, therefore the roots come in complex-conjugate pairs, that gives the desired conclusion.

Let's count the zeros in the right half plane with the formula $$\frac{1}{2\pi i}\int \frac{f'(z)}{f(z)}\, dz =\frac{1}{2\pi i}\int\frac{6z^5+192}{z^6+192z+640}\, dz ;$$ here, the integration is over a large semicircle from $-iR$ to $iR$, and then back along the imaginary axis. The semicircle makes a contribution of $\approx 6/2=3$ because $f'/f\simeq 6/z$ there. To evaluate the contribution coming from the segment on the imaginary axis, we observe that $f(it)\not= 0$, so we can take a holomorphic logarithm on a neighborhood of the path from $iR$ to $-iR$, and then $$\frac{1}{2\pi i} \int_{iR}^{-iR} \frac{f'(z)}{f(z)}\, dz = \frac{1}{2\pi i}(\log f(iR) -\log f(-iR)) .$$ Now just keep track of how $f(it)=-t^6 +640 + 192it$ moves as we vary $t$ from $R$ to $-R$: $f(iR)$ starts out in the third quadrant, then crosses the positive real axis (when $t=0$) and then moves to $\overline{f(iR)}$ in the lower half plane. This shows that $\log f(iR)-\log f(-iR) = -i\varphi$, $\pi<\varphi<2\pi$.

This gives the (approximate) bounds $3-1+o(1)\le N\le 3-1/2+o(1)$ on the number of zeros in the right half plane. In other words, there are exactly two zeros, and since the zeros come in complex conjugate pairs (and there are obviously no zeros on the axes), all claims follow now.