# zeros of a polynomial

Given $P(z)=z^6+6z+10$, find how many roots are in each quadrant

I have already seen that $P(z)$ has six different roots, and that none of them are real or of the form $ki$, $k\in \Bbb R$. Since the coefficients of $P(z)$ are real, then it's roots are 3 conjugate pairs, and therefore the possibilities are limited to two.

Defining $f(z)=z^6$, $g(z)=z^6+6z+10$ and applying Rouche in
$|z|<2$, $|f(z)-g(z)|<|f(z)|$
I get that all roots are contained in the disc, but how do I go about splitting them into the four quadrants? From what I know about this polynomial, finding zeros in one quadrant is enough to determine the rest, but I'm stuck trying to apply rouche to a quarter circle (basically I'm having trouble finding a suitable f and g)

• Try doing a quarter circle contour centered at the origin and employ Rouche. Rouche is not limited to circular contours. – Cameron Williams May 26 '14 at 0:52
• I forgot to mentiont that this is exactly what I'm having trouble with, choosing the appropriate f and g to apply rouche to a quarter circle – notacat May 26 '14 at 1:20
• You may use Routh-Hurwitz theorem. Also, even if it's not needed, Rouché's theorem applied to $z^6$ and $6z+10$ shows there is no root with $|z|<1$, so all roots lie in the annulus $1<|z|<2$. – Jean-Claude Arbaut May 26 '14 at 20:12
• thanks for all the help, seeing 3 different methods to tackle this is great! – notacat May 28 '14 at 11:02

Rather than using Rouché's theorem we can approach this as a perturbation problem. To wit, consider the polynomial

$$P_a(z) = z^6 + az + 10.$$

One can show that if $a \neq 0$ then $P_a$ has no zeros on the imaginary axis and if $0 \leq a < 6 \cdot 2^{5/6}$ then $P_a$ has no zeros on the real axis either.

If $a = 0$ then $P_a$ has six simple zeros -- one in each quadrant and one at each point $z = \pm 10^{1/6} i$.

By the inverse function theorem these zeros are analytic functions of $a$ for $a$ small enough. If $z = z(a)$ is one of these zeros then

$$z^6 + az + 10 = 0$$

and, differentiating with respect to $a$,

$$6z^5 z' + z + az' = 0,$$

so that

$$z' = - \frac{z}{a + 6z^5}.$$

If we consider the zero with $z(0) = 10^{1/6} i$ then

$$z'(0) = - \frac{1}{6 \cdot 10^{2/3}}.$$

Consequently, the zero of $P_a$ located at $z = 10^{1/6} i$ when $a = 0$ moves into the left half-plane as $a$ increases past zero. We remarked earlier that $P_a$ has no purely imaginary zeros if $a \neq 0$, so this zero must lie in the left half-plane for all $a > 0$.

Since the coefficients of the polynomial are real, the same is true for the zero located at $z = -10^{1/6} i$ when $a = 0$.

Each of the zeros starting in the four quadrants when $a = 0$ must remain in their quadrant for all $0 \leq a < 6 \cdot 2^{5/6}$, and once the zeros on the imaginary axis fall into the quadrants in the left half-plane they must remain there as well.

Taking $a = 6$ (and noting that $6 < 6\cdot 2^{5/6}$) we conclude that $P$ has two zeros in each of the quadrants II and III and one zero in each of the quadrants I and IV.

• really liked how it was a completely different approach, thanks! – notacat May 28 '14 at 11:03