Quantity of zeros on a ring - Complex Analysis

[SqueeSpleen]

Determine the quantitiy of zeroes of the function:
$f(z)=z^{4}-8z+10$
a) Inside the circle $| z | < 1$
b) Inside the ring $1 \leq | z | < 2$a)
$f(z)=(z^{4}-8z)+10=g(z)+h(z)$
As $|h(z)| \geq |g(z)| \forall z : | z | = 1$
Then by Rouche's Theorem the number of zeros of the function inside the circle is the same than $h(z)$ (0 zeroes).
b) My idea was to calculate the number of zeroes in $| z | < 2$ and substract the number of zeroes in $| z | < 1$.
But I can't find a pair of functions to use Roche's Theorem.
Any hint?
I ploted here:
http://www.wolframalpha.com/input/?i=z^4-8z%2B10
And I'm starting to suspect that the statement is wrong, the roots are very close to the circle and I guess that this is what make the functions so hard to find.
But I may be wrong and perhaps it's a easy way to decompose $f(z)$

 

[mfb]

(a) is correct.

And I'm starting to suspect that the statement is wrong

Which statement?

As all roots appear in pairs, you just have to distinguish between three cases for the numbers of roots inside. It looks like a clever contour integral can help, but I'm not sure how exactly.

 

[SqueeSpleen]

The function, because b) is very hard =/ the modulus of a pair of roots is 1.993.

 

[mfb]

WolframAlpha gives solutions that are well separated from 2 in their magnitude.

 

[SqueeSpleen]

Sorry... I forget to take the square root, so it isn't 1.993, its 1.412.
Thank you, I'll continue trying (I tried 5-6 decompositions until I used wolframalpha and I thought it was way harder than it's).

 

[SqueeSpleen]

My professor said that the exercise was miswritten, it was meant to be easier to use Rouche's theorem than others method.

 

[jackmell]

My professor said that the exercise was miswritten, it was meant to be easier to use Rouche's theorem than others method.

Very good then. I hate it when I can't do these. Ask your professor to let us to this one instead (same problem):

$$z^7-5z^3+12$$

 

[SqueeSpleen]

Around the same contour this one is easy, because $| 2^{7} | > | 40 | + | 12 | \geq | 5z^{3} + 12 |$ (When $| z | = 2$)
(Triangular inequality).
Then you have 7 zeroes here (As $z^{7}$ has 7).
As you have:
$| 12 | > | 5 | + | 1 | \geq | 5z^{3} + z^{7} |$ (When $| z | = 1$) you have 0 zeroes in the circle of radius 1.
Then all the zeroes are in the ring of outer radius 2 and inner radius 1.
(If the zeroes are n in the circle of radius 1 and m in the circle of radius 2, in the ring they're m-n).

 

[jackmell]

Around the same contour this one is easy, because $| 2^{7} | > | 40 | + | 12 | \geq | 5z^{3} + 12 |$ (When $| z | = 2$)
(Triangular inequality).
Then you have 7 zeroes here (As $z^{7}$ has 7).
As you have:
$| 12 | > | 5 | + | 1 | \geq | 5z^{3} + z^{7} |$ (When $| z | = 1$) you have 0 zeroes in the circle of radius 1.
Then all the zeroes are in the ring of outer radius 2 and inner radius 1.
(If the zeroes are n in the circle of radius 1 and m in the circle of radius 2, in the ring they're m-n).

Very good. If the objective of this exercise is to expose and teach you Rouche's Theorem, then going through just this easy one is much better in my opinion then a lesson in frustration with one we can't do. :)