Suppose we have complex mapping f:X→Yf : X \rightarrow Y, where X,Y⊂CX, Y \subset \mathbb{C}. How do we visualize it? As, we can't visualize four dimension data in a single plot, we split it into two, one for the domain and one for the codomain.

For example, consider the unit cycle centered at 22

S1≐{z∈C:∣z−2∣<1}. S_1 \doteq \{ z \in \mathbb{C} : | z - 2 | < 1 \}.

Let f(z)≐2z−1f(z) \doteq \frac{2}{z - 1}. Then S1S_1 is mapped to the set

S2≐{z∈C:Re(z)>1}. S_2 \doteq \{ z \in \mathbb{C} : \text{Re}(z) > 1 \}.

This is shown by the picture below.

We verify this algebraically.

Proof 1. Let w=2z−1w = \frac{2}{z - 1}. As w≠0w \neq 0, we have

z−1=2w  ⟹  z=2w+1. {\small z - 1 = \frac{2}{w} \implies z = \frac{2}{w} + 1. }

Hence,

∣z−2∣<1  ⟹  ∣2w−1∣<1  ⟹  ∣2−ww∣2<1  ⟹  ∣2−w∣2<∣w∣2  ⟹  ∥(2−x,y)∥22<∥(x,y)∥22let w=(x,y)  ⟹  4−4x<0  ⟹  1<x. \begin{aligned} \small |z - 2 | < 1 &\implies \left| \frac{2}{w} - 1 \right| < 1 \\ &\implies \left| \frac{2 - w}{w} \right|^2 < 1 \\ &\implies \left| 2 - w \right|^2 < |w|^2 \\ &\implies \Vert (2 - x, y) \Vert_2^2 < \Vert (x, y) \Vert_2^2 &\text{let } w = (x, y) \\ &\implies 4 - 4 x < 0 \\ &\implies 1 < x. \end{aligned}

□\square

Proof 2. Denote the border of S1S_1 as z=2+eiθz = 2 + e^{i \theta}, where θ∈[−π,π)\theta \in [-\pi, \pi). Then

w=2z−1=21+eiθ=21+eiθ⋅e−iθ/2e−iθ/2=2e−iθ/2e−iθ/2+eiθ/2=2⋅cos⁡(θ/2)+isin⁡(θ/2)2cos⁡(θ/2)=1+itan⁡(θ/2). \begin{aligned} w &= \frac{2}{z - 1} \\ &= \frac{2}{1 + e^{i \theta } } \\ &= \frac{2}{1 + e^{i \theta } } \cdot \frac{e^{- i\theta / 2 } }{ e^{ - i \theta / 2 } } \\ &= \frac{2 e^{ - i \theta / 2} }{ e^{ - i \theta / 2 } + e^{i \theta / 2} } \\ &= 2 \cdot \frac{ \cos (\theta / 2) + i \sin (\theta / 2) }{ 2 \cos (\theta / 2) } \\ &= 1 + i \tan (\theta / 2). \end{aligned}

□\square