9
$\begingroup$

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)

$\endgroup$
  • 2
    $\begingroup$ Try doing a quarter circle contour centered at the origin and employ Rouche. Rouche is not limited to circular contours. $\endgroup$ – Cameron Williams May 26 '14 at 0:52
  • $\begingroup$ 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 $\endgroup$ – notacat May 26 '14 at 1:20
  • 1
    $\begingroup$ 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 $1<|z|<2$. $\endgroup$ – Jean-Claude Arbaut May 26 '14 at 20:12
  • $\begingroup$ thanks for all the help, seeing 3 different methods to tackle this is great! $\endgroup$ – notacat May 28 '14 at 11:02
3
$\begingroup$

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 $z = \pm 10^{1/6} i$.

enter image description here

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 $a = 0$.

enter image description here

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.

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

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.