Using the binomial theorem to prove that $z^n$ is continuous in $\mathbb{C}$

Question

How can I prove that $f(z)=z^n$ is a continuous function in $\mathbb{C}$ for every $n$ in $\mathbb{N}$ using the binomial theorem?
I did $h=z-z_0$ and got $z^n = (h + z_0)^n$, and that equals the binomial theorem.
But does it prove that $z^n$ is continuous? I thought it was obvious because $z^n$ is a polynomial function and those are always continuous, so I don't really know why the need to use the binomial theorem.

Answer 1

By the binomial theorem:

$$(z+h)^n = \sum_{t=0}^n {n \choose t} h^t z^{n-t} = z^n + \sum_{t=1}^n {n \choose t} h^t z^{n-1} = z^n + h\left(\sum_{t=1}^n {n \choose t} h^{t-1} z^{n-t}\right)$$

It is clear now that as $h \to 0$, $f(z+h)=(z+h)^n\to z^n = f(z)$. So $f$ is continuous.

Answer 2

First, start with your definition of continuity. I prefer the definition that $f(z)$ is continuous if at each point $z_{0}$ in $\mathbb{C}$, we have that $\lim_{z \to z_{0}} f(z) = f(z_{0})$.

In other words, if we choose a particular $z_{0}$, and let $z = h + z_{0}$, then we want to express $f(z)$ as $$f(z_{0} + h) = f(z_{0}) + g(h)$$ for some function $g(h)$, where $\lim_{h \to 0} g(h)= 0$. Note that $g$ depends on the choice of $z_{0}$.

The point here is that the binomial theorem allows us to find a convenient expression for $g(h)$. This should help with proving that the limit is zero.