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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?

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That's good. Now just observe that you can take $\delta<1$, so $|z+2|<4$ and therefore, for $|z-1|<\delta=\min(\varepsilon/4,1)$ you have $$ |(z^2+z+1)-3|=|z-1|\,|z+2|<\frac{\varepsilon}{4}\cdot 4=\varepsilon $$

Note that, if $|z-1|<1$, you have $$ |z+2|=|(z-1)+3|\le |z-1|+3<4 $$


On the other hand, the function $g(z)=z^2+z+1$ is everywhere continuous and coincides with $f$ on $\mathbb{C}\setminus\{1\}$. Therefore $$ \lim_{z\to 1}f(z)=\lim_{z\to 1}g(z)=g(1)=3 $$ by definition of limit and continuity.

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  • $\begingroup$ Thank you so much for your help. That was elegant. $\endgroup$ – bloodtypebpos Feb 21 '16 at 17:36
  • $\begingroup$ @bloodtypebpos More generally, if $g$ is continuous on some open set $U$ and differentiable at $z_0\in U$, the function $f(z)=(g(z)-g(z_0))/(z-z_0)$, for $z\ne z_0$, and $f(z_0)=g'(z_0)$ is continuous at $z_0$ (and so in the whole set $U$). $\endgroup$ – egreg Feb 21 '16 at 17:40
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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.

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