Application of Rouche's Theorem HELP

[arithmuggle]

Question: Find the number of zeroes of the equation $$z^7 - 2z^5 + 6z^3 - z+1 = 0$$, in the unit disk.


We have Rouche's theorem which says that if f(z) and g(z) are two functions analytic in a neighborhood of the closed unit disk and if |f(z) - g(z)| < |f(z)| for all z on the boundary of the disk, then f and g have the same number of zeroes in the disk.

Now there are easier problems which I can do just fine. But here is what I try. Now first off I know from here-say and mathematica that there are three roots inside the unit disk. So naturally I could use g(z) is either "6z^3" or "-2z^5 + 6z^3" since both have three zeroes (counting multiplicity) INSIDE the disk. So here is all i can see:

using the identity $$\right|a+b\right| \le \left|a\right|-\left|b\right|$$over and over again I get:
|f(z)| >= 1 immediately not very helpful...
from below I can see, using $$g(z) = -2z^5 + 6z^3$$,
$$\left|f(z)-g(z)\right| = \left|z^7 - z + 1\right| < 3$$ where i deliberately use strict < if you think about the geometry of this triangle inequality application...

ok so I can't get my |f(z)-g(z)| to be smaller than |f(z)|... any help?

 

[exalted_shmo]

Try using the alternate (equivalent formulation) of Rouche's theorem
$$|f(z) - g(z)| < |f(z)| + |g(z)|$$

using $$f(z) = z^7 - 2z^5 + 6z^3 - z + 1$$ and

Spoiler

$$g(z) = 6z^3$$

.

Because then

Spoiler

$$|f(z)-g(z)| = |z^7 - 2z^5 - z + 1|$$
$$|f(z)-g(z)| \leq 5 < 6 \leq |6z^3|$$
$$|f(z)-g(z)| < |z^7 - 2z^5 + 6z^3 - z + 1| + |6z^3| = |f(z)| + |g(z)|$$

.

There is a very similar example to your problem in "Handbook of complex variables" by Krantz on page 74.

 

[arithmuggle]

Ahh thank you so much. I see that version of the theorem in Ahlfors. Although, I don't see why I would ever use the version of the theorem I had originally quoted since clearly having that extra "$$+ \left| g(z) \right|$$" makes these inequalities a heck of a lot easier.