Prove that $(\frac{1}{n} +z )^2$ doesn't converge uniformly on $\mathbb{C}$. The problem states:

Prove or disprove that $\left(\frac{1}{n}+z\right)^2$ does not converge uniformly on $\mathbb{C}$.

My attempt
The first thing I did was to show that $f_n = \frac{1}{n} +z$ converges uniformly on $\mathbb{C}$ to $z$. Now I want to see if the sequence of squares $(f_n^2)_{n\in\mathbb{N}}$ is converging also or not.
I presume it doesn't so, let $\epsilon = \frac{1}{2}$.
$$\left|(\frac{1}{n} +z)^2- z^2\right| = \left|\frac{1}{n^2} + 2\frac{z}{n} +z^2 - z^2\right| + \left|\frac{1}{n^2}+2\frac{z}{n}\right| \leq \frac{1}{n^2} + 2\frac{|z|}{n}$$
If we let $z=n$, the last expression will be
$$\frac{1}{n^2} + 2 >\frac{1}{2},$$
Hence, it is not uniformly convergent.
 A: Your proof is not entirely correct, but your idea is correct. What is correct? that if you find a subsequence $(z_n)_n$ such that its always $\frac{1}{2}$ or more away from its limit, then it is not uniformly convergent. However, you proved that
$$\frac{1}{n^2}+2>\frac{1}{2}$$
which doesn't tell you anything about $(\frac{1}{n}+z)^2-z^2$ since it is upper-bounded (instead of lower-bounded) by $\frac{1}{n^2}+2>\frac{1}{2}$
Since $$f_n\to z\implies f_n^2\to z^2$$
By replacing $z=n$ on your first term we get $$\left|\left(\frac{1}{n}+n\right)^2-n^2\right| = \left|\frac{1}{n}+2\right| = 2+\frac{1}{n}>\frac{1}{2}$$
So it is not uniformly convergent.
A: That is not correct. You have proved that $|f_n(n)-f(n)|\leqslant2+\frac1{n^2}$ (where $f(z)=z^2$). So what? This doesn't prove that you do not have $|f_n(n)-f(n)|\leqslant\frac12$, in spite of the fact that $\frac1{n^2}+2>\frac12$.
However, note that $|f_n(n)-f(n)|\color{red}=2+\frac1{n^2}$. It follows from this that, indeed, the convergence is not uniform.
