Let $m,n\in\mathbb{N}$, I have to prove that the polynomial $P(z)=1+z+\dfrac{z^2}{2!}+...+\dfrac{z^m}{m!}+3z^n$ has exactly $n$ roots in the $D(0,1)$.
My attempt: Let's call $Q(z)=3z^n$, $R(z)=1+z+\dfrac{z^2}{2!}+...+\dfrac{z^m}{m!}$. Then $P(z)=Q(z)+R(z)$. We are going to apply the Rouché Theorem. Let $z\in\partial D(0,1)$, then $|z|=1$ and $|Re(z)|\leq|z|=1$.
Then, $|R(z)|=|\sum_{k=1}^m\dfrac{z^k}{k!}|\leq|\sum_{k=1}^\infty\dfrac{z^k}{k!}|=|e^z|=e^{|Re(z)|}\leq e < 3=|3z^n|=|Q(z)|$
So, by the Rouché Theorem, $P(z)=Q(z)+R(z)$ and $Q(z)$ has the same zeros in $D(0,1)$. As the zeros of $Q(z)=3z^n$ are $z=0$ with multiplicity $n$, $P(z)$ has $n$ zeros in $D(0,1)$.
My problem is that I don't know how to prove the first inequality. I've tried some ways without results. Some help?
Your inequalities are not correct.
The correct argument is $|R(z)| \leq \sum\limits_{k=1}^{m} \frac {|z|^{k}} {k!}\leq \sum\limits_{k=1}^{\infty} \frac {|z|^{k}} {k!} =e^{|z|} = e <|3z^{n}|=|Q(z)|$.