How can the number of zeros of a complex function in a given domain be proven?

[jjr]

Homework Statement

Let $D={z : |z| <1}$. How many zeros (counted according to multiplicty) does the function $f(z)=2z^4-2z^3+2z^2-2z+9$ have in $D$? Prove that you answer is correct.

Homework Equations

3. The Attempt at a Solution [/B]
The function has no zeros in $D$, which can be seen quite easily because $2z^4-2z^3+2z^2-2z>-9$ when $|z|<1$. I am having some trouble proving this though. A suggestion I have is to use the inequality $|2z^4-2z^3+2z^2-2z|<8$. I can get as far as the next step $|z^4-z^3+z^2-z| < 4$, but I am not sure how to go on from this point.

Any hints or tips would be greatly appreciated!J

 

[S.G. Janssens]

At this point I would invoke Rouché's theorem.

 

[mfb]

Triangle inequality?

 

[jjr]

Thanks! So if I understand correctly:

$|z^4-z^3+z^2-z| \leq |z^4| + |z^3| + |z^2| + |z| \leq 4$ because $|z^n|<1$ if $|z|<1$, where $n \in {1,2,3,4}$. And because $8<9$ it's proven.J

 

[S.G. Janssens]

Yes, exactly! (Minor detail: I would write "because $|z^n| \le 1$ for $z \in \partial D$ and $n = 1,2,3,4$" since in the formulation of the theorem the estimate is actually required to hold only on the boundary $\partial D$ of $D$.)