Attractor/Examples/Origin under Complex Square Function

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Example of Attractor

Consider the complex function $f: \C \to \C$ defined as:

$\forall z \in \C: \map f z = z^2$

Then the origin $\tuple {0, 0}$ of the Argand plane is an attractor.


Proof

Consider the set of points in the open (complex) disk with the center at the origin $\tuple {0, 0}$ and radius $1$ which is the set:

$\map B {0, 1} = \set {z \in \C: \cmod z < 1}$


All of these points can be described in exponential form:

\(\ds z\) \(=\) \(\ds r e^{i \theta}\) Definition of Complex Number/Polar Form
\(\ds \) \(=\) \(\ds e^{\ln r} e^{i \theta}\) Exponential of Natural Logarithm
\(\ds \) \(=\) \(\ds e^{\ln r + i \theta}\) Exponential of Sum

Since $\cmod z < 1$ and from Modulus of Exponential is Exponential of Real Part, then $\cmod r < 1$.

Applying our squaring complex function to any of the points in this set, we obtain:

\(\ds \map f z\) \(=\) \(\ds \paren {e^{\ln r + i \theta} }^2\)
\(\ds \) \(=\) \(\ds \paren {e^{2 \ln r + 2 i \theta} }\) Exponential of Product

Applying our squaring complex function to any of the points in this set $n$ times, we obtain:

\(\ds \map f {\map f {\cdots \map f z} }\) \(=\) \(\ds \paren {e^{\ln r + i \theta} }^{2^n}\)
\(\ds \) \(=\) \(\ds \paren {e^{2^n \ln r + 2^n i \theta} }\) Exponential of Product

Since $\cmod r < 1$ and from Natural Logarithm of 1 is 0 and Logarithm is Strictly Increasing, we know $\ln r < 0$.

Therefore, we have:

$\ds \lim_{n \mathop \to \infty} 2^n \ln r = -\infty$

And from Exponential Tends to Zero and Infinity, we have:

$\ds \lim_{n \mathop \to \infty} \paren {e^{2^n \ln r + 2^n i \theta} } = 0$


$\blacksquare$

Sources

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