Given that the complex plane has two dimensions, if $\lbrace a_n \rbrace$ converges to $L\in\mathbb{C}$, we say that for every $\epsilon >0$, $\exists\Delta>0$ such that $\forall n\geq\Delta$, $a_n\in B_\epsilon(L)$
Hence, if we wanted to show that $$\lim_{n\to\infty}z^n=0,~~~~~|z|<1$$ we could apply our "adapted" epsilon-delta procedure. Hence, let $L=0$, and let $a_n=z^n$. Then, for $\lbrace a_n \rbrace$ to converge to $L$, there must exist pairs of $\epsilon$ and $\Delta$ greater than $0$ such that $$n\geq\Delta ~~\implies~~|a_n|<\epsilon$$and since $a_n=z^n$, we have $$|z^n|<\epsilon$$ given that $|z|<1$, if we were to take logarithms of both sides, we arrive at $$\log|z^n|>\log\epsilon$$ and since $n>0$,
$$n\log|z|>\log\epsilon$$ $$n>\frac{\log\epsilon}{\log|z|}$$ $\implies$ for every $\epsilon$, if we let $$\Delta=\frac{\log\epsilon}{\log|z|}$$then every $n>\Delta \implies |a_n|<\epsilon$, which concludes the proof.
Question in short is, is this method correct? I've been rather sloppy in terms of inequality signs, if anyone spots an error just let me know. Any other responses are appreciated, thank you.
This is wrong from the start. You talk about pairs of $\varepsilon$ and $\Delta$ greater than $0$. What you are supposed to prove is that for any $\varepsilon>0$, there is some $\Delta>0$ such that $n>\Delta\implies|z^n|<\varepsilon$.
Furthermore, $\log$ is an increasing function. Therefore, $|z|^n<\varepsilon\iff\log\bigl(|z|^n\bigr)<\log\varepsilon$. You reverted the inequality sign.