Does anyone know how to show that all the roots of the following complex polynomial lie in the annulus 1/2 < |z| < 1?

9z⁵ + 5z -3 = 0


Johnny

Andre

Alternative is to apply Rouche's theorem. If you don't know about Rouche's theorem, then here is a more elementary way:

(1) Show that there are no roots with |z| < = 1/2.
[Hint: Suppose there is a root. Write the equation in the form 3=-5z-9z⁵ . Taking modulus of both sides and apply triangle inequality to the right hand side to get a contradiction.]

(2) Similarly, show that there are no roots with |z| > = 1.
[Hint: Rewrite the equation in the form 3-5z=9z⁵ . Then take modulus of both side and use triangle inequality to get a contradiction.]

(3) Using (1) and (2), deduce all roots of the equation must lie in the annulus.

Kerwin

Another related question I have is finding the number of zeros in each quadrant. I have used the argument principle (ap) to determine one zero in each of quadrant II and III, but I'm having some difficulty applying it to quadrant I and IV because there is a root on the positive real axis. So my main problem is trying to draw a contour curve so that it would go around that real root before applying ap. Any suggestions on how to do this?

If you have proved that there is exactly 1 root in quadrant II (i.e. top left) and exactly 1 root in quadrant III (i.e. bottom left) [note: I have not checked this] then you are done. You know there is exactly 1 real root, that leaves two roots left (it?s a quintic equation). But the coefficients are real, therefore if z is a root, so is z^* .

So one root is in quadrant I (top right) and one in quadrant IV (bottom right).

Do you agree?

Andre

I would agree if there is exactly 1 real root. But the problem I had was proving that it is actually so. So I was thinking about drawing a contour around that real root, but it makes the whole problem messy.

By the way, could you remind me why if z is a root then its conjugate is also a root as well?

Firstly, if

f(x) = 9x⁵ + 5x - 3

where x is real, then we observe that

f(x) -> -infinity as x -> -infinity
f(x) -> infinity as x -> infinity

so f(x) crosses the x axis, so there is at least 1 real root. But if you look at

f'(x) = 45x⁴ + 5

you also see that its gradient is always positive ==> there is only 1 root.

Secondly, suppose that z satisfies

9z⁵ + 5z - 3 = 0

Take complex conjugate of both sides:

(9z⁵ + 5z - 3)^* = (0)^* = 0

But (a+b)^* = a^* + b^* [for any a,b -real or complex] therefore

(9z⁵ )^* + (5z)^* - (3)^* = 0
(9z⁵ )^* + (5z)^* -3 = 0

But (ab)^* = a^* b^* [for any a,b -real or complex] therefore (since powers are repeated multiplication)

(9)^* (z⁵ )^* + (5)^* (z)^* -3 = 0
9(z^* )⁵ + 5(z^* ) -3 = 0

In other words, z^* satisfies exactly the same equation. Note that we relied in several places on the coefficients being real .

Andre

To show that p(z) has only one real root, and that it lies within the annulus, construct a Sturm chain

f0 = 9z⁵ + 5x - 3
f1 = 45z⁴ + 5
f2 = 3 - 4z
f3 = -4925/256

and find that there is a real root in (-1,1), since the number sign changes of the chain evaluated at -1 and 1 differ by one.

Thanks Andre, that was crystal clear.
Thanks Ben for an alternate way in proving the uniqueness of the root.
And thanks Kerwin for the hints to my orignial question (I think I got it now).


Johnny