I'm looking at some notes from my previous complex variables course and I need help verifying some things about the convergence of $$ \prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right) $$ on compact subsets of $\mathbb{C}$. My professor shows its convergence by using the property of infinite products which states that $\prod_{n=1}^{\infty}(1-a_n)$ converges if and only if $\sum_{n=1}^{\infty}\log(1-a_n)$ converges, given $a_n\neq 1$ for each $n$. I have a few issues I would like to work out with this:
$\mathbf{\frac{z^2}{n^2}=1}$ when $\mathbf{z=\pm n}$, which one would think makes the property inapplicable.
I believe I have maneuvered around this hurdle because if $z$ belongs to an arbitrary compact subset of $\mathbb{C}$, then $\frac{z^2}{n^2}=1$ for only finitely many $n$, there exists an $N\in\mathbb{N}$ such that $\frac{z^2}{n^2}\neq 1$ for all $n\geq N$, and it suffices to check convergence of $\sum_{n=N}^{\infty}\log\left(1-\frac{z^2}{n^2}\right)$. The only thing that concerns me is that this point wasn't mentioned in the notes whatsoever.
Convergence $\mathbf{\sum_{n=1}^{\infty}\log\left(1-\frac{z^2}{n^2}\right)}$ is shown by noticing $\mathbf{\sum_{n=1}^{\infty}\left|\frac{z^2}{n^2}\right|}$ converges uniformly on compact subsets of $\mathbf{\mathbb{C}}$ (through the Weierstrass M-test, for example).
I've thought this through but I don't see it at all. I was thinking it might be because $\log\left(1-\frac{z^2}{n^2}\right)$, could be thought of as a function of $\frac{z^2}{n^2}$, but this cannot hold always since $\sqrt{\frac{1}{n^2}}$ is a function of $\frac{1}{n^2}$ but the sum of these terms clearly does not converge.
Any help on these points is greatly appreciated. Thanks in advance.
For your first question: If $z=\pm n$ then $1-\tfrac{z^2}{n^2}=0$, so the entire product equals $0$. To see that the product converges we remove this term from the product, and the argument below serves just as well to show that the product converges.
Your second question does require some effort. I don't see how to show that $\sum_{n=1}^{\infty}\log\left(1+\tfrac{z^2}{n^2}\right)$ converges, but I'm able to prove the product converges uniformly on compact sets using a lovely little lemma on infinite products. I've chopped the proof up into three parts:
$$\ $$
1. Uniform convergence of $\sum_{n=1}^{\infty}\left|\tfrac{z^2}{n^2}\right|$ on compact subsets of $\Bbb{C}$
Let $B\subset\Bbb{C}$ compact and $M\in\Bbb{R}$ such that $|z|\leq M$ holds for all $z\in B$. The series $\sum_{n=1}^{\infty}\tfrac{1}{n^2}$ can be seen to converge, say to $L$,${}^{\dagger}$ by noting that it is strictly positive and increasing, and satisfies $$\sum_{n=1}^k\frac{1}{n^2}<1+\sum_{n=2}^k\frac{1}{n(n-1)}=1+\sum_{n=2}^k\left(\frac{1}{n-1}-\frac{1}{n}\right)=1+\left(1-\frac{1}{k}\right)=2-\frac{1}{k},$$ for all $k\geq1$. It follows that $\sum_{n=1}^{\infty}\left|\tfrac{z^2}{n^2}\right|$ converges uniformly to $|z|^2L$ on $B$, as for all $z\in B$ we have $$\left|\sum_{n=1}^k\left|\frac{z^2}{n^2}\right|-|z|^2L\right|=|z|^2\left|\sum_{n=1}^k\frac{1}{n^2}-L\right|\leq M^2\cdot\left|\sum_{n=1}^k\frac{1}{n^2}-L\right|.$$ This indeed means the series converges uniformly, because the series $\sum_{n=1}^{\infty}\tfrac{1}{n^2}$ converges to $L$.
$$\ $$
2. An often useful lemma on infinite products
Lemma: If a sequence $(\alpha_n)_{n=1}^{\infty}$ satisfies $\sum_{n=1}^{\infty}|1-\alpha_n|\leq\frac{1}{2}$, then for all $m\geq1$ we have $$\left|1-\prod_{n=1}^m\alpha_n\right|\leq2\sum_{n=1}^m|1-\alpha_n|.$$
Proof: We proceed by induction. The base case $m=1$ is obvious, so suppose the inequality holds for some $m\geq1$. From the induction hypothesis we get \begin{align} \left|\prod_{n=1}^m\alpha_n\right|\leq1+\left|1-\prod_{n=1}^m\alpha_n\right|\leq1+2\sum_{n=1}^m|1-\alpha_n|\leq2.\tag{1} \end{align} Then it is a matter of some clever rewriting to see that \begin{eqnarray*} \left|1-\prod_{n=1}^{m+1}\alpha_n\right| &=&\left|1-\prod_{n=1}^m\alpha_n +\prod_{n=1}^m\alpha_n -\prod_{n=1}^{m+1}\alpha_n\right|\\ &\leq&\left|1-\prod_{n=1}^m\alpha_n\right| +\left|\prod_{n=1}^m\alpha_n\right| \cdot\left|1-\alpha_{m+1}\right|\\ &\leq&2\sum_{n=1}^m\left|1-\alpha_n\right|+2\left|1-\alpha_{m+1}\right| =2\sum_{n=1}^{m+1}\left|1-\alpha_n\right|.\tag*{$\Box$} \end{eqnarray*} Note that this shows in particular that $\left|1-\prod_{n=1}^m\alpha_n\right|\leq1$ holds for all $m\geq1$.
$$\ $$
3. Actually showing that the product converges
Now there exists $N\in\Bbb{N}$ such that $\sum_{n=N}^{\infty}\left|\frac{z^2}{n^2}\right|\leq\frac{1}{2}$ holds for all $z\in B$; simply take $N$ such that $$\sum_{n=N}^{\infty}\frac{1}{n^2}\leq\frac{1}{2M^2}.$$ Hence we can apply the lemma to the sequence $(p_n(z))_{n=1}^{\infty}$ given by $p_n(z):=1-\tfrac{z^2}{(N+n)^2}$, because $$\sum_{n=1}^{\infty}|1-p_n(z)|=\sum_{n=1}^{\infty}\left|\frac{z^2}{(N+n)^2}\right|=\sum_{n=N+1}^{\infty}\left|\frac{z^2}{n^2}\right|\leq\frac{1}{2}.$$ Note that if $z=\pm k$ then $N>|k|$, so $p_n(z)$ is nonzero for all $n\geq1$. Setting $P_m(z):=\prod_{n=1}^mp_n(z)$ we can show the sequence $(P_m(z))_{m=1}^{\infty}$ converges uniformly on $B$. The lemma together with inequality (1) shows that for integers $a>b\geq1$ we have \begin{eqnarray*} \left|P_a(z)-P_b(z)\right| &=&\left|\prod_{n=1}^bp_n(z)\right| \cdot\left|1-\prod_{n=b+1}^ap_n(z)\right|\\ &<&2\cdot2\sum_{n=b+1}^a|1-p_n(z)| =4\sum_{n=b+1}^a\left|\frac{z^2}{n^2}\right|\\ &\leq&4M^2\sum_{n=b+1}^a\frac{1}{n^2}. \end{eqnarray*} This shows that $|P_a(z)-P_b(z)|\ \rightarrow\ 0$ as $a,b\ \rightarrow\ \infty$, so $(P_m(z))_{m=1}^{\infty}$ converges uniformly on $B$, to $$\lim_{m\to\infty}P_m(z)=\lim_{m\to\infty}\prod_{n=N+1}^{N+m}\left(1-\frac{z^2}{n^2}\right)=\prod_{n=N+1}^{\infty}\left(1-\frac{z^2}{n^2}\right).$$ It follows immediately that we also have uniform convergence on $B$ for the product $$\prod_{n=1}^{\infty}\left(1+\frac{z^2}{n^2}\right)=\prod_{n=1}^N\left(1+\frac{z^2}{n^2}\right)\cdot\prod_{n=N+1}^{\infty}\left(1+\frac{z^2}{n^2}\right).$$
$$\ $$
$\dagger$ In fact the series can be shown to converge to $L=\tfrac{\pi^2}{6}$, this is the Basel problem.
For completeness sake I include my original answer below. It only shows convergence on the disk $|z|<(2\pi^2)^{-1}$, but it also shows a few interesting connections.
Using the power series of $\operatorname{Log}(1-x)$ we find that
$$\operatorname{Log}\left(1-\frac{z^2}{n^2}\right)=\sum_{k=1}^{\infty}\frac{\left(\tfrac{z^2}{n^2}\right)^k}{k}=\sum_{k=1}^{\infty}\frac{z^{2k}}{kn^{2k}},$$
so interchanging the order of summation (why is this allowed?) yields $$\sum_{n=1}^{\infty}\operatorname{Log}\left(1-\frac{z^2}{n^2}\right)=\sum_{k=1}^{\infty}\frac{z^{2k}}{k}\sum_{n=1}^{\infty}\frac{1}{n^{2k}},$$ where now the inner sum (over $n$) is an instance of the Riemann zeta function, which for the positive even integers just so happens to converge to $$\sum_{n=1}^{\infty}\frac{1}{n^{2k}}=\zeta(2k)=(-1)^{k+1}B_{2k}\frac{(2\pi)^{2k}}{2(2k)!},$$ where $B_m$ is a Bernoulli number. Hence we can simplify the above to $$\sum_{k=1}^{\infty}\frac{z^{2k}}{k}\cdot(-1)^{k+1}B_{2k}\frac{(2\pi)^{2k}}{2(2k)!}=-\sum_{k=1}^{\infty}\frac{B_{2k}}{2(2k)!}\frac{(-2\pi^2z^2)^k}{k}.$$ Now it suffice to note that $0<B_{2k}<2(2k)!$ for sufficiently large $k$, which can be seen using Stirling's approximations for $B_{2k}$ and $(2k)!$. Now the series above converges whenever the series $$-\sum_{k=1}^{\infty}\frac{(-2\pi^2z^2)^k}{k}=\operatorname{Log}(1+2\pi^2z^2),$$ converges, which is the case for $|z|<(2\pi^2)^{-1}$ as can be seen by the ratio test, for example.