# 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).

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook