Branch Point/Examples/Cube Root of z-a

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Example of Branch Point

Let $f: \C \to \C$ be the complex function defined as:

$\forall z \in \C: \map f z = \paren {z - a}^{1/3}$

for some $a \in \C$.

Then $a$ is a branch point of $f$.


Proof

We need to show that:

$f$ has exactly one value at $a$ itself.
$f$ has more than one value at one or more points in every neighborhood of $a$


We have that:

$f$ has exactly one value at $a$ itself:
$\map f a = \paren {a - a}^{1/3} = 0$


To demonstrate the second part, we recall that the $\epsilon$-neighborhood of $a$ is defined as:

$\map {N_\epsilon} a := \set {z \in \C: \cmod {z - a} < \epsilon}$

Therefore, we need to show that $\forall \epsilon: \map f {\epsilon + a}$ will have more than one value.

From Cube Roots of Unity we have:

\(\ds e^{0 i \pi / 3}\) \(=\) \(\ds 1\)
\(\ds e^{2 i \pi / 3}\) \(=\) \(\ds -\frac 1 2 + \frac {i \sqrt 3} 2\)
\(\ds e^{4 i \pi / 3}\) \(=\) \(\ds -\frac 1 2 - \frac {i \sqrt 3} 2\)


We can rewrite $\epsilon$ as $e^{\map \ln \epsilon}$ thus:

$\map f {\epsilon + a} = \paren {\paren {\epsilon + a} - a}^{1/3} = e^{\dfrac 1 3 \map \ln \epsilon} \times \paren {1}$
$\map f {\epsilon + a} = \paren {\paren {\epsilon + a} - a}^{1/3} = e^{\dfrac 1 3 \map \ln \epsilon} \times \paren {-\dfrac 1 2 + \dfrac {i \sqrt 3} 2 }$
$\map f {\epsilon + a} = \paren {\paren {\epsilon + a} - a}^{1/3} = e^{\dfrac 1 3 \map \ln \epsilon} \times \paren {-\dfrac 1 2 - \dfrac {i \sqrt 3} 2 }$


Hence $f$ has more than one value (three in this example) at one or more points in every neighborhood of $a$.

$\blacksquare$


Sources

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