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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?

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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.

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