How to show the function $f\colon z\mapsto\lvert z\rvert^2$ is continuous on the entire complex plane?

Question

I am required to show this using the epsilon-delta definition of continuity and I'm having trouble applying what I know about real functions to complex ones. I've let $z_0$ be a complex number in $\mathbb{C}$ and then considered all points in the open-disc $D(z_0,1):=\lbrace z:\lvert z-z_0\rvert<1\rbrace$. I know that $\lvert f(z)-f(z_0)\rvert={\large\lvert}\lvert z\rvert^2-\lvert z_0\rvert^2{\large\lvert}$ but I'm not sure how I can relate this to $\lvert z-z_0\rvert<1$. Is difference of two squares the method required?

Answer 1

I think you can try to rewrite the original function $f(z) = |z|^{2}$ as $f(z) = z z^{*}$ where $z^{*}$ is the complex conjugacy.

Answer 2

Let $z = x+\mathrm{i}y$ and $\delta = \delta_x + \mathrm{i} \delta_y$. Then \begin{align*} |z+\delta| &= |(x+\delta_x) + \mathrm{i}(y +\delta_y)| \\ &\leq |x+\delta_x| + |y +\delta_y| \text{.} \end{align*}

With a little triangle inequality manipulation, you can get the inequality you need.

Answer 3

For any sequence $(z_n)_{n \in \mathbb{N}}$ with $z_n \stackrel{n \to \infty}{\longrightarrow} z_0$ you have $$\begin{eqnarray*} \left| |z_n|^2 - |z_0|^2\right|& = & \left| (|z_n| - |z_0|)(|z_n| + |z_0|)\right| \\ & = & \left| |z_n| - |z_0|\right|\cdot\left(|z_n| + |z_0|\right) \\ & \leq & \left| z_n - z_0\right|\cdot \left(|z_n| + |z_0|\right) \\ & \stackrel{n \to \infty}{\longrightarrow} & 0\cdot 2|z_0| = 0 \end{eqnarray*}$$

So $f$ is continuous on $\mathbb{C}$.