All roots of a specific polynomial lie inside the unit disc I am currently working on the following problem:
Let $m,n \in \mathbb{N}$ and $p: \mathbb{C} \to \mathbb{C},~ z \mapsto z^{m+1}-z^m+(\frac{1}{4n})^m$.
Show that if $p(z)=0$ then $|z|<1$, i.e. that all roots of the polynomial $p$ lie inside the open unit disc.
I first approached the problem by trying to use results from complex analysis:

*

*The theorem of Gershgorin yields an upper bound on the absolute value of the roots, which is strictly larger than 1

*Rouché's theorem does not seem to be applicable here in a suitable way, since the sum of any two coefficients is larger than the remaining coefficient.

Next, I tried to argue by contradiciton:
Assume that there exists a $z\in \mathbb{C}$ with $|z|\geq 1$ such that $p(z)=0$, so
$z^{m+1}-z^m+(\frac{1}{4n})^m = 0$ or equivalently $z^m(z-1) = -(\frac{1}{4n})^m$. Then for the absolute value we have
\begin{aligned}
 |z^m(z-1)| & =  (\frac{1}{4n})^m \\
|z^m||(z-1)| & =  (\frac{1}{4n})^m \\
|z-1| & = (\frac{1}{4n|z|})^m \leq (\frac{1}{4n})^m 
\end{aligned}
Thus, it holds $z \in B_{\leq (\frac{1}{4n})^m}(1)$ and $z \notin B_{\leq 1}(0)$. The polynomial $\zeta^m(\zeta-1)$ has a single real root at 1, so I think it should be possible to show that the root $z$ of $p$ close to 1 also is real. This would yield a contradiciton becuase then there exists an $\varepsilon > 0$ such that $z = 1+ \varepsilon$, but then we would have
$(1+\varepsilon)^m(1+\varepsilon-1) = (1+\varepsilon)^m \varepsilon \overset{!}{=} -(\frac{1}{4n})^m$.
I was not capable to show that the root is real, my only idea is to argue about the complex argument of $z^m(z-1)$, which would be 0 because $-(\frac{1}{4n})^m$ is real.
Any ideas are highly appreciated!
 A: If $m=1$ then the roots can be computed explicitly as $\frac 12 \pm \frac 12 \sqrt{1-\frac 1n}$, therefore I'll assume $m\ge 2$ in the following.
Remark: The following approach is motivated by experiments with Wolfram Alpha which show that for larger $m$, $p$ has one real zero very close to (but less than) one, and the remaining zeros are roughly the solutions of $z^m=a^m$.
We have $p(z) = z^{m+1}-z^m+a^m$ where $a=1/(4n)$ is a real number in the range $(0, 1/4]$.
First consider the disk $B_{1/2}(0)$ and compare $p$ with $q(z) = z^m-a^m$: For $|z|=1/2$ is
$$
|p(z)-q(z)| = |z^{m+1}| = \frac{1}{2^{m+1}} < \frac{1}{2^{m}} - \frac{1}{4^{m}} \le |z|^m - |a|^m \le |z^m-a^m| = |q(z)|
$$
and Rouché's theorem shows that $p$ has exactly $m$ zeros in $B_{1/2}(0)$.
Now consider the real function
$$
 f(x) = x^{m+1}-x^m + a^m
$$
and show that $f(1/2) < 0 < f(1)$, so that $f$ has a zero in the interval $(1/2, 1)$. That is the remaining root of $p$, and it is also inside the unit disk.
A: One can give a direct proof as follows; by $w=1/z$ the problem is equivalent to showing that $cw^{m+1}-w+1=0$ has all roots $|w|>1$ where $0 <c \le \frac{1}{4^m}$
But now assume $|w| \le 1$ is a root and note that then $|1-w| \le c$ and if we let $w=re^{i\theta}$ so in particular $|\theta| \le \pi/4$ as otherwise $\Re (1-w) \ge 1-\sqrt 2/2>1/4$,  one can then easily see that $|1-e^{i\theta}| \le |1-w| \le c$
(for example from the isosceles triangle $OAB, O=0, A=1, B=e^{i\theta}$ where $w=C \in |OB|$, then in the triangle $ABC$ we have $|AC| \ge |AB|$ since the opposite angles satisfy same inequality as $AC$ is inside the angle $OAB=OBA=CBA$)
But now this means $2|\sin \theta/2| \le c$ and using $|\sin \theta/2| \ge |\theta|/\pi$ one gets $|\theta| \le \pi c /2$ which means $(m+1)|\theta| \le (m+1)c \pi /2$ so $|\arg w^{m+1}| \le \frac{(m+1)\pi}{2 \times 4^m}\le \pi/4$ which gives $\Re cw^{m+1} >0$ so $\Re (cw^{m+1}-w+1) \ge \Re cw^{m+1}>0$ Contradiction!
