Determine the number of solutions of equation $27z^{11}-18z+10=0$ where $|z|<1$ using Rouché's Theorem I'm trying to get the number of zeroes of function $F(z)=27z^{11}-18z+10=0$. So, I have to choose $f(z)$ and $g(z)$ such that $F(z)=f(z)+g(z)$ and $|f(z)|>|g(z)|$ on |z|=1.
I tried choosing following functions:
$f(z)=27z^{11}, g(z)=-18z+10$
When $|z|=1:$
$|27z^{11}|=27; |10-18z|\le10+18=28$. Fail $(27<28)$.
$f(z)=27z^{11}-18z, g(z)=10$
When $|z|=1:$
$|27z^{11}-18z|\ge||27z^{11}|-|18z||=27-18=9; |10|=10$; Fail $(9<10)$.
$f(z)=10, g(z)=27z^{11}-18z$
When $|z|=1:$
$|10|=10;|27z^{11}-18z|\le|27z^{11}|+|18z|=45$; Fail $(10<45)$.
So, I can't prove that $|f(z)|>|g(z)|$.
What I am doing wrong?
 A: Ok, I've been dealing with this problem for a while, and according the answer given in the book (excercise 435, page 132. Solution given is 11), and due that I have already found some other typos across the book before, my bet is that the equation should have been:
$$27z^{11}-18z+1=0$$
And the reason is that this equation has his 11 solutions in $|z|<1$ if the independent coefficient is less than 9. Say:
$$27z^{11}-18z+a=0$$
Then, let $g(z)=27z^{11}-18z+a$ and $f(z)=27z^{11}$, where $f(z)$ has its 11 roots in $|z|<1$, then:
$$|f(z)-g(z)|=|27z^{11}-(27z^{11}-18z+a)|=|18z-a| \leq 18|z|+|a| =18+|a|$$
And:
$$|f(z)|=|27z^{11}|=27|z|^{11}=27$$
So the inequality $|f-g|<|f|$ holds iff:
$$18+|a|<27$$
Therefore $$|a|<9$$

I have also tried to prove (unsuccessfully) that the original equation
$$27z^{11}-18z+10=0$$
has only 10 roots in $|z|<1$ by using a $f(z)=Az^{10}$, so $|f(z)|=|A|$ and then:
$$
\begin{equation}
    \begin{split}
    |f(z)-g(z)|&=|Az^{10}-(27z^{11}-18z+10)|=|Az^{10}-27z^{11}+18z-10|\\
    &\leq |A|+27+18+10=|A|+55
    \end{split}
\end{equation}
$$
But the $|f-g|<|f|$ cannot be hold because it would imply that:
$$|A|+55<|A|$$
which I'm quite sure that cannot be.
So my guessing for this is that there must be a polynomial $h(z)=a_nz^n+a_{n-1}z^{n-1}+\dots+a_1z+a_0$ with its $n$ roots in $|z|<1$ such that, multiplied by $g(z)=27z^{11}-18z+10$ we can obtain a new polynomial $fh$ with 10 out of its $n+11$ roots in $|z|<1$. But according to the level of the book, maybe this would be too much work for an exercise that should be a lot more simple accordingly to the level of the others exercises.
