# Proving the continuity of a complex valued function

So I am given the following function

$$f(z)\begin{cases} z & |z|\leq 1 \\ |z|^2 & |z| > 1 \end{cases}$$ and I want to find all points of continuity. I know that for $$|z| < 1$$ and $$|z| > 1$$ that I am fine; the only issue is possibly at $$|z| = 1.$$ So we want to do the usual thing and show that for any $$z_0$$ such that $$|z_0| = 1$$ that $$\lim_{z \to z_0}f(z) = f(z_0).$$ But since $$|z_0| = 1,$$ we know that $$f(z_0) = z_0.$$ Similarily if we chose any sequence lying inside the closed unit disc, then we will be fine since the function is the identity here. So we are only concerned with a sequence $$z_n$$ such that $$|z_n| > 1.$$ So suppose that $$z_n$$ is a complex valued sequence which converges to $$z_0.$$ We see that $$f(z_n) = |z_n|^2 = (\sqrt{x_n^2 + y_n^2})^2 = x_n^2 + y_n^2.$$ I am not sure where to go here other than using the fact that if a sequence $$z_n$$ converges to $$z_0,$$ then $$|z_n|$$ converges to $$|z_0|.$$ Could we then conclude that there is possibly a problem here since $$f(z_n) = 1,$$ but $$z_0$$ is not necessarily 1? For example, we can consider the following sequence $$z_n = \bigg(\frac{\sqrt{2}}{2} + \frac{1}{n}\bigg) + i\bigg(\frac{\sqrt{2}}{2} + \frac{1}{n}\bigg)$$ which converges to $$z_0 = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2},$$ which has modulus 1, but $$f(z_n)$$ converges to $$\frac{\sqrt{2}}{2}^2 + \frac{\sqrt{2}}{2}^2$$ = 1. And then conclude that the function is not continuous here?

You have computed a sequence $$z_n$$ that converges to $$z_0$$ and shown that $$f(z_n)$$ does not converge to $$f(z_0)$$. So you have shown that $$f$$ is not continuous at $$z_0$$.
You can use this same principle to check continuity at any other point on the unit circle. Let $$z_0$$ be any fixed point with $$|z_0|=1$$, let $$z_n$$ be a sequence that converges to $$z_0$$ such that $$|z_n|>1$$ for all $$n$$. Then $$\lim f(z_n)= \lim |z_n|^2 = 1$$. So $$f$$ is continuous at $$z_0$$ if and only if $$f(z_0)=1$$, that is if $$z_0=1$$, for all other $$z_0$$ with $$|z_0|=1$$, $$f$$ is not continuous.