Show that limlimits_(z→ a)z^n=a^n for n∈mathbb N , i.e. that…
Show that $\ \lim\limits_{z\to a}z^n=a^n\ $ for $n\in\mathbb N$, i.e. that $f(z)=z^n$ is continuous.
Explanation of Method
Factor $z^n-a^n$ using the standard geometric-sum identity, bound the resulting sum uniformly once $z$ is confined to a bounded neighborhood of $a$, and produce an explicit $\varepsilon$-$\delta$ pair. An independent induction proof via the product rule is included as a cross-check.
Step 1 · Factor the difference
$$ z^n-a^n=(z-a)\sum_{k=0}^{n-1}z^{n-1-k}a^k $$
Step 2 · Bound $|z|$ near $a$
Suppose $|z-a|<1$. Then $|z|\le|a|+|z-a|<|a|+1=:R$.
Step 3 · Bound the sum
Each term satisfies $|z|^{n-1-k}|a|^k<R^{n-1-k}R^k=R^{n-1}$, so $$ \left|\sum_{k=0}^{n-1}z^{n-1-k}a^k\right|\le\sum_{k=0}^{n-1}|z|^{n-1-k}|a|^k<nR^{n-1} $$
Step 4 · Combine and choose $\delta$
$$ |z^n-a^n|<|z-a|\cdot nR^{n-1} $$
Given $\varepsilon>0$, choose $$ \delta=\min\left\{1,\ \frac\varepsilon{nR^{n-1}}\right\},\qquad R=|a|+1 $$ Then $|z-a|<\delta\implies|z^n-a^n|<\varepsilon$, proving $\lim_{z\to a}z^n=a^n$.
Alternative · induction via the product rule
Base case $n=1$: $f(z)=z$ is trivially continuous. Inductive step: if $z^k$ is continuous, then $z^{k+1}=z\cdot z^k$ is a product of continuous functions, hence $$ \lim_{z\to a}z^{k+1}=\lim_{z\to a}z\cdot\lim_{z\to a}z^k=a\cdot a^k=a^{k+1} $$ so continuity propagates to every $n\in\mathbb N$ by induction.
Result
Both proofs generalize immediately: the $\varepsilon$-$\delta$ argument is the template for showing any polynomial is continuous, while the induction argument is the template for showing any finite product of continuous functions is continuous.