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 $g(z) = -2z^5 + 6z^3$,
$\left|f(z)-g(z)\right| = \left|z^7 - z + 1\right| &lt; 3$ 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)| &lt; |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 &lt; 6 \leq |6z^3|
|f(z)-g(z)| &lt; |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.