Let $n\geq 3$. Show that the polynomial $z^n+nz-1$ has $n$ zeros in $D(0,R)$, where $$R=1+\left(\frac{2}{n-1}\right)^{1/2}.$$
I was hoping to use Induction and Rouche's Theorem. For the base case (n=3) I was easily able to show that the polynomial has $3$ zeros inside $|z|=R$. It's the base case that's giving me trouble. I assume that, for $k\geq 3$, the polynomial has exactly $k$ zeros inside $|z|=1+\left(\frac{2}{k-1}\right)^{1/2}$. Then I want to look at what happens with the polynomial $z^{k+1}+(k+1)z-1$. Letting $f(z)=z^{k+1}$, $g(z)=(k+1)z-1$, and considering $R=1+\left(\frac{2}{k}\right)^{1/2}$, I can't figure out the moduli of $f$ and $g$ and compare them to verify that I can use Rouche's Theorem.
Is induction the wrong way to go here?
On $|z|=R$ we have that $|z^n|=R^n$ and
$$|nz-1|\leq nR+1=n+n\left(\frac{2}{n-1}\right)^{1/2}+1$$
If we prove that
then $$|nz-1|<|z^n|$$ on $|z|=R$ and by Rouche $z^n+nz-1$ and $z^n$ have the same number of zeros in the interior of $|z|=R$.
It is the inequality above the one you would like to prove for $n\geq3$.
We can expand using the Binomial Theorem $$\begin{align}\left(1+\left(\frac{2}{n-1}\right)^{1/2}\right)^n&=1+n\left(\frac{2}{n-1}\right)^{1/2}+\underbrace{\frac{n(n-1)}{2}\left(\frac{2}{n-1}\right)}\limits_{n}+\text{ other positive terms}\\&>1+n\left(\frac{2}{n-1}\right)^{1/2}+n\end{align}$$
Therefore the inequality is satisfied.