Linear and Quadratic function on $\mathbb{C}$
1. Linear function on $\mathbb{C}$
$f(z) = \alpha z + \beta$ $(\alpha, \beta \in \mathbb{C})$
$= \alpha (z + \frac{\beta}{\alpha})$
$= \frac{\alpha}{\left|\alpha \right| } \cdot \left|\alpha \right| (z + \frac{\beta}{\alpha})$
$+ \frac{\beta}{\alpha}$ : translate
$\left|\alpha \right| $ : times
$\frac{\alpha}{\left|\alpha \right| }$ : CCW rotation
Ex
$w = (1+i)z + 2$
Since $1 + i = \sqrt{2} (\cos \frac{\pi}{4} + \sin \frac{\pi}{4})$,
$\Rightarrow$ $\sqrt{2}$ times + $\frac{\pi}{4}$ CCW rotation + translated by $(2, 0)$
2. Quadratic function on $\mathbb{C}$
$f(z) = \alpha z^{2} + \beta z + \gamma$ $(\alpha, \beta, \gamma \in \mathbb{C})$
$= \alpha(z + \frac{\beta}{2 \alpha})^{2} + \gamma_{1} - \frac{\beta^{2}}{4 \alpha}$
$g(z) = z + \frac{\beta}{2 \alpha}$ : translate
$h(z) = z^{2}$
$k(z) = \alpha z + \gamma - \frac{\beta^{2}}{4 \alpha}$ : linear function
$\therefore$ $f(z) = k \circ h \circ g(z)$
2-1. $z^{2}$ on $\mathbb{C}$
$f(z) = z^{2}$
$z = x + iy \mapsto z^{2} = (x^{2} - y^{2}) + i(2xy)$
$F : \mathbb{R}^{2} \to \mathbb{R}^{2}$ by
$(x, y) \mapsto (x^{2} - y^{2}, 2xy)$
Let $u = x^{2} - y^{2}$, $v = 2xy$
$\Rightarrow$
$x^{2} = \dfrac{\sqrt{u^{2} + v^{2}} + u}{2} $
$y^{2} = \dfrac{\sqrt{u^{2} + v^{2}} - u}{2} $
Ex 1 $y = mx$
$z^{2} : (x, mx) \mapsto (x^{2}(1 - m^{2}), 2mx^{2})$
$\left\{\begin{matrix}
u = x^{2}(1-m^{2}) \\
v = 2mx^{2} \end{matrix}\right.$ $\Rightarrow$ $v = \dfrac{2m}{1-m^{2}}u$
$\mathrm{arg}(z) = \theta$ or $\mathrm{arg}(z) = \theta + \pi$ $\Rightarrow$ $\mathrm{arg}(z^{2}) = 2\theta$
Ex 2 $y = c$
$z^{2} : (x, c) \mapsto (x^{2} - c^{2}, 2cx)$
$\left\{\begin{matrix}
u = x^{2} - c^{2} \\
v = 2cx \end{matrix}\right.$ $\Rightarrow$ $u = \dfrac{v^{2}}{4c^{2}} - c^{2}$
Ex 3 $x = c$
$z^{2} : (c, y) \mapsto (c^{2} - y^{2}, 2cy)$
$\left\{\begin{matrix}
u = c^{2} - y^{2} \\
v = 2cy \end{matrix}\right.$ $\Rightarrow$ $u = -\dfrac{v^{2}}{4c^{2}} + c^{2}$
Ex 4
$\left\{z \; | \; 0 \leq \mathrm{arg}(z) \leq \frac{\pi}{2} \right\}$ $\Rightarrow$ $\left\{z \; | \; 0 \leq \mathrm{arg}(z^{2}) \leq \pi \right\}$
Ex 5
$\left\{z \; | \; \left|z \right| \leq 1 \right\}$ $\Rightarrow$ $\left\{z \; | \; \left|z \right| \leq 1 \right\}$