Show that the four of the five roots of

1. Feb 3, 2012

Jamin2112

Show that the four of the five roots of ....

1. The problem statement, all variables and given/known data

Show that four of the five roots of z5 + 15z + 1 = 0 belong to the annulus {z: 3/2 < |z| < 2}.

2. Relevant equations

Argument Principle (presumably)

3. The attempt at a solution

Since f(z) = z5 + 15z + 1 is entire, it has no poles. Thus, if N1 is the number of zeros inside C2(0), N2 is the number of zeroes inside C3/2(0), and N is the number of zeroes inside the annulus {z: 3/2 < |z| < 2},

N = N2 - N1 = 1/(2πi) [ ∫C2(0)f'(z)/f(z) dz - ∫C3/2(0)f'(z)/f(z) dz ]
Right? Or is there an easier way?

EDIT: Ah, shoot! That assumes that |z|=3/2 and |z|=2 have no zeroes.
EDIT 2: And also that no zeroes are that |z|>2.

What's the strategy here?

2. Feb 4, 2012

morphism

Re: Show that the four of the five roots of ....

It's better to use Rouche's theorem here.

3. Feb 4, 2012

Jamin2112

Re: Show that the four of the five roots of ....

Right, and with it I can show that 5 roots are in the range |z| < 2 and 1 root is in the range |z| < 3/2. Then I just need to show that no roots are on the circles |z| = 2 or |z| = 3/2. Eh?

4. Feb 4, 2012

morphism

Re: Show that the four of the five roots of ....

So you know that there are 4 roots in 3/2<=|z|<2 (you don't need to worry about |z|=2 - think about what Rouche's theorem says to see why).

To exclude the possibility of there being a root on the circle |z|=3/2, you can repeat the same Rouche argument with 3/2+epsilon in place of 3/2 to show that there's one root in |z|<3/2+epsilon (for sufficiently small epsilon).