Show that the function $f$ is continuous at the given point
$$f(z) =\begin{cases}
\dfrac{z^3 – 1}{z-1} & \text{if } |z| \neq 1 \\[6pt]
3 & \text{if } |z| = 1 \\
\end{cases}$$ $z_0 = 1$
I have the following definition for the Limit of a Complex Function:
Suppose that a complex function $f$ is defined in a deleted neighborhood
of $z_0$ and suppose that $L$ is a complex number. The limit of $f$ as $z$
tends to $z_0$ exists and is equal to $L$, written as $\lim\limits_{z \to z_0} f(z) = L$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $\vert f(z) – L \vert < \epsilon $ whenever $0 < \vert z – z_0 \vert < \delta$.
My intuition is to start with $\vert f(z) – L \vert < \epsilon$ and manipulate this until the expression on the left looks like $\vert z-z_0 \vert$. From there, I figured I would have identified an appropriate value for my $\delta$ and used it to work forward for my proof. However, I got a little stuck with the manipulation. I can see that for $z = z_0 = 1$ that $\vert z_0 \vert = 1$ and so $f(z_0) = 3$ immediately. When I work in the neighborhood of $z_0$ for $\vert z \vert \neq 1$ that I do approach $3$. I've decided to try an epsilon/delta proof for the condition that $vert z \vert \neq 1$. Here's what I've done so far:
$$
\begin{align}
f(z) & = \frac{z^3 – 1}{z-1} \\[6pt]
& = \frac{z^3 – 1^3}{z-1} \\[6pt]
& = \frac{(z-1)(z^2 + z + 1^2)}{z-1} \\[6pt]
& = z^2 + z + 1
\end{align}
$$
From here, I can plug this alternate form of $f(z)$ into $\vert f(z) – L \vert < \epsilon$ and continue to manipulate:
$$
\begin{align}
\varepsilon & > \vert (z^2 + z + 1) – (3) \vert \\[6pt]
& = \vert z^2 + z – 2 \vert \\[6pt]
& = \vert (z-1)(z+2) \vert
\end{align}
$$
This is where I'm stuck. I don't think that I can just divide $\epsilon$ by $\vert z+2 \vert$ because that would make it depend on the variable $z$ which defeats setting a fixed value for $\delta$. What am I supposed to do here?
Best Answer
Observe that $$\vert (z-1)(z-1+3) \vert \leq \vert (z-1)^2 \vert + 3\vert (z-1) \vert \leq \delta^2+3\delta.$$ Therefore consider the quadratic equation $$\delta^2+3\delta-\epsilon=0$$ Solving this should give you a positive $\delta$ in terms of $\epsilon$ which you desire.