Proving that this function converges uniformly.

Question

I was wondering if there is an easier way to show that this sequence of functions converges uniformly. Also I am almost $100 \% $ sure that my reasoning is not a proof, so yeah, help me please (there must be other way).

$$ f_{n}(z)=\frac{e^{-n|z| }}{n} $$ With $z\in \mathbb{C} $

I thought that since this sequence tends to zero no matter what is the absolute value of $z$ . Being this limit the zero function in $\mathbb{C}$. The only way that maybe break the apparent uniform convergence is by choosing $$f(z^*)$$ such that $$|z^*| = \ln\frac{1}{n^{\frac{1}{n}}}$$ so that $$f(z^*)=\frac{1}{1} $$ but in order to get this i would have to choose a natural $n$ such that the absolute value of $z^*$ is of course positive, and this would be a $n$ such that $$ 0<n^{\frac{1}{n}}< 1$$ so there is no $n$ that satisfies this condition.

Answer 1

It converges uniformly to $f(z)=0$ because given $\epsilon>0$, let $N>1/\epsilon$, so $\forall n>N$

$$|f_n(z)-f(z)|=|f_n(z)|=\left|\frac{e^{-n|z| }}{n}\right|=\frac{e^{-n|z| }}{n}\leq\frac{e^{-n0 }}{n}=1/n\leq 1/N<\epsilon$$

Answer 2

Since $e^{-n|z|} \le 1$ for all $z\in \Bbb C$ and $n\in \Bbb N$, then $$\max_{z\in \Bbb C} |f_n(z)| \le \frac{1}{n}\quad (n = 1,2,3,\ldots)$$ which implies $(f_n)$ converges uniformly to $0$.