2.4 參考解答

(a) $\rm{II}$ . $w=f_1(z)=|z|^{\frac{1}{2}}e^{i\frac{\text{Arg}\space (z)}{2}}=r^{\frac{1}{2}}e^{i\frac{\theta}{2}}=r^{\frac{1}{2}}(\cos{\frac{\theta}{2}}+i\sin{\frac{\theta}{2}})$ , $-\pi< \theta \leq \pi$
$x<0$ , $y>0$
Let $z=re^{i\theta}$ , $r>0$ , $\frac{\pi}{2}<\theta<\pi$
$\therefore w=f_1(z)=\rho e^{i\phi}$ , $\phi=\frac{\theta}{2}$ , $\rho=r^{\frac{1}{2}}>0$
$\frac{\pi}{2}<\theta<\pi \Rightarrow \frac{\pi}{4}<\frac{\theta}{2}<\frac{\pi}{2}, \frac{\pi}{4}<\phi<\frac{\pi}{2}$
Hence, the sector $\rho =r^{\frac{1}{2}}>0$ & $\frac{\pi}{4}<\phi<\frac{\pi}{2}$ .
(b) $\rm{II}$ . $w=f_2(z)=|z|^{\frac{1}{2}}e^{i\frac{\text{Arg}+2\space (z)+2\pi}{2}}=r^{\frac{1}{2}}e^{i\frac{\theta+2\pi}{2}}$
Let $z=re^{i\theta}$ , $r>0$ , $\frac{\pi}{2}<\theta<\pi$
$w=f_2(z)=\rho e^{i\phi}$ , $\rho=r^{\frac{1}{2}}>0$ , $\phi=\frac{\theta+2\pi}{2}$
$\frac{5\pi}{2}<\theta+2\pi<3\pi \Rightarrow \frac{5\pi}{4}<\frac{\theta+2\pi}{2}<\frac{3\pi}{2}, \frac{5\pi}{4}<\phi<\frac{3\pi}{2}$
Hence, the sector $\rho =r^{\frac{1}{2}}>0$ & $\frac{5\pi}{4}<\phi<\frac{3\pi}{2}$ .

5. $f_1(z)=|z|^{\frac{1}{3}}e^{i\frac{\text{Arg}\space (z)}{3}}=r^{\frac{1}{3}}(\cos{\frac{\theta}{3}}+i\sin{\frac{\theta}{3}})$ , $|z|=r$ , $\theta=\text{Arg}\space (z)$

(a) $\forall \space z=re^{i\theta} \in \mathbb{C}$ , $r>0$ , $\theta=\text{Arg} \space (z)$ ,

then $[f_1(z)]^3=(|z|^{\frac{1}{3}}e^{i\frac{\text{Arg}\space (z)}{3}})^3=|z|e^{i\text{Arg}\space (z)}=z$

$\therefore f_1$ is a branch of the cube root function.

(b) { $z=re^{i\theta}|r>0$ , $-\pi< \theta \leq \pi$ }

Let $f_1(z)=\rho e^{i\phi}=|z|^{\frac{1}{3}}e^{i\frac{\text{Arg}\space (z)}{3}}$

$\rho =|z|^{\frac{1}{3}}>0$ , $-\pi<\theta \leq \pi \Rightarrow -\frac{\pi}{3}<\frac{\text{Arg}\space (z)}{3} \leq \frac{\pi}{3}$

$\text{image}(f_1)=$ { $\rho e^{i\phi}\space|\space \rho >0$ , $-\frac{\pi}{3}<\phi \leq \frac{\pi}{3}$ }

$\Rightarrow f_1$ on $\mathbb{C}$ \ $\{(x,0)\space|\space x \leq 0\}$

$\lim_{(r,\theta) \rightarrow (r_0,\pi)} f_1(re^{i\theta})=\lim_{(r,\theta) \rightarrow (r_0,\pi)} r^{\frac{1}{3}}(\cos{\frac{\theta}{3}}+i\sin{\frac{\theta}{3}})=r_0^{\frac{1}{3}}(\frac{1}{2}+\frac{\sqrt{3}}{2}i)$

$\lim_{(r,\theta) \rightarrow (r_0,-\pi)} f_1(re^{i\theta})=\lim_{(r,\theta) \rightarrow (r_0,-\pi)} r^{\frac{1}{3}}(\cos{\frac{\theta}{3}}+i\sin{\frac{\theta}{3}})=r_0^{\frac{1}{3}}(\frac{1}{2}-\frac{\sqrt{3}}{2}i)$

$f_2(z)=r^{\frac{1}{3}}\cos{(\frac{\theta+2\pi}{3}})+ir^{\frac{1}{3}}\sin{(\frac{\theta+2\pi}{3}})$ , where $r>0$ , $-\pi<\theta \leq \pi$
(a) $\forall \space z=re^{i\theta} \in \mathbb{C}$ , $r>0$ , $\theta=\text{Arg}\space (z)$
$(f_2(z))^3=(r^{\frac{1}{3}}e^{i\frac{\theta+2\pi}{3}})^3=re^{i(\theta+2\pi)}=re^{i\theta}=z$
$\therefore f_2$ is a branch of the cube root function.
(b) $\{ z=re^{i\theta}\space|\space r>0, -\pi<\theta \leq \pi \}$
Let $f_2(z)=\rho e^{i\phi}$ & $f_2(z)=r^{\frac{1}{3}}e^{i\frac{\theta+2\pi}{3}}$
$\therefore \rho =r^{\frac{1}{3}}$ & $\rho >0$
$-\pi<\theta \leq \pi \Rightarrow \pi <\theta+2\pi \leq 3\pi$
$\Rightarrow \frac{\pi}{3}< \frac{\theta+2\pi}{3} \leq \pi$ (i.e. $\frac{\pi}{3}<\phi \leq \pi$ )
$\therefore \text{image}\space (f_2)=\{\rho e^{i\phi}\space|\space \rho>0, \frac{\pi}{3}<\phi \leq \pi\}$
(c) f_2 is continuous everywhere except $z=0$ and the point on the negative $x$ -axis.
(d) $0$ is the branch point.

9. (a) $w=f(z)=z^{\frac{1}{3}}$ , $z=re^{i\theta}$ , $-\pi<\theta \leq \pi$

$f_1(z)=r^{\frac{1}{3}}e^{i\frac{\theta}{3}}$ , $f_2(z)=r^{\frac{1}{3}}e^{i\frac{\theta+2\pi}{3}}$ , $f_3(z)=r^{\frac{1}{3}}e^{i\frac{\theta+4\pi}{3}}$

Let $D_1=\{re^{i\theta}: r>0, -\pi<\theta \leq \pi\}$

$D_2=\{re^{i\theta}: r>0, \pi<\theta \leq 3\pi\}$

$D_3=\{ re^{i\theta}: r>0, 3\pi<\theta \leq 5\pi \}$

從 $x$ 軸的負數軸部分剪開， $D_3$ 在 $D_2$ 上方， $D_2$ 在 $D_1$ 上方，

而 $D_1$ 的上半平面接在 $D_2$ 的下半平面 ( $\theta=\pi$ )

$D_2$ 的上半平面接在 $D_3$ 的下半平面 ( $\theta=3\pi$ )

$D_3$ 的上半平面接在 $D_1$ 的下半平面 ( $\theta=-\pi$ )

( $\theta=5\pi$ )

其這3層的domain分別使 $f(z)=z^{\frac{1}{3}}$ 為1-1的function。