Finding Zeroes of a Complex Polynomial Inside |z| < 1

[Gauss M.D.]

Homework Statement

Find the number of zeroes of

p(z) = z^5 + 10z - 1

inside |z| < 1

Homework Equations

The Attempt at a Solution

Let f(z) = z^5
g(z) = 10z-1

On |z| = 1:

10|z| - 1< |10z-1| < 10|z| + 1 (is this true...?)

9 < |g(z)| < 11

|f(z)| = 1

So |g| > |f|, so f+g should have the same number of zeroes as f, which is five.

That's incorrect obviously. What am I doing wrong?

 

[pasmith]

Homework Statement

Find the number of zeroes of

p(z) = z^5 + 10z - 1

inside |z| < 1

Homework Equations

The Attempt at a Solution

Let f(z) = z^5
g(z) = 10z-1

On |z| = 1:

10|z| - 1< |10z-1| < 10|z| + 1 (is this true...?)

Set z = Re^{i\theta}. Then
|10z - 1|^2 = (10Re^{i\theta} - 1)(10Re^{-i\theta} - 1) = 100R^2 + 1 + 20R\cos\theta
so that
<br /> 100R^2 - 20R + 1 \leq |10z - 1|^2 \leq 100R^2 + 20R + 1<br />
or
<br /> (10R - 1)^2 \leq |10z - 1|^2 \leq (10R + 1)^2<br />
so yes, it is true that 10|z| - 1 \leq |10z-1| \leq 10|z| + 1.

9 < |g(z)| < 11

|f(z)| = 1

So |g| > |f|, so f+g should have the same number of zeroes as f, which is five.

That's incorrect obviously. What am I doing wrong?

You haven't applied Rouche's Theorem correctly: if |g| &gt; |f| on |z| = 1 then g and f + g have the same number of zeroes in |z| &lt; 1.

 

[Gauss M.D.]

But if |g| is between 9 and 11 and f is always 1, then f < g on |z| = 1. What's incorrect?

 

[pasmith]

But if |g| is between 9 and 11 and f is always 1, then f < g on |z| = 1. What's incorrect?

What's incorrect is your conclusion that f and f +g have the same number of zeroes. Rouche's Theorem is actually telling you that g and f + g have the same number of zeroes.