2.5 參考解答

  1. D43(−2i3)=D_{\frac{4}{3}}(-\frac{2i}{3})={z:∣z+2i3∣<43z: |z+\frac{2i}{3}|<\frac{4}{3}}, f(z)=1zf(z)=\frac{1}{z}

    w=f(z)=1z⇔z=f−1(w)=1w=uu2+v2+i−vu2+v2w=f(z)=\frac{1}{z} \Leftrightarrow z=f^{-1}(w)=\frac{1}{w}=\frac{u}{u^2+v^2}+i\frac{-v}{u^2+v^2}

    ∴x=uu2+v2\therefore x=\frac{u}{u^2+v^2}, y=−vu2+v2y=\frac{-v}{u^2+v^2}

    ∣z+2i3∣<43⇔(uu2+v2)2+(23+−vu2+v2)2<(43)2|z+\frac{2i}{3}|<\frac{4}{3} \Leftrightarrow (\frac{u}{u^2+v^2})^2+(\frac{2}{3}+\frac{-v}{u^2+v^2})^2<(\frac{4}{3})^2

    ⇔u2(u2+v2)2+49−43vu2+v2+v2(u2+v2)2<169\Leftrightarrow \frac{u^2}{(u^2+v^2)^2}+\frac{4}{9}-\frac{4}{3}\frac{v}{u^2+v^2}+\frac{v^2}{(u^2+v^2)^2}<\frac{16}{9}

    ⇔1u2+v2−4v3(u2+v2)−43<0\Leftrightarrow \frac{1}{u^2+v^2}-\frac{4v}{3(u^2+v^2)}-\frac{4}{3}<0

    ⇔3−4v−4(u2+v2)<0\Leftrightarrow 3-4v-4(u^2+v^2)<0

    ⇔4u2+4v2+4v−3>0\Leftrightarrow 4u^2+4v^2+4v-3>0

    ⇔(u−0)2+(v+12)2>1\Leftrightarrow (u-0)^2+(v+\frac{1}{2})^2>1

    which is exterior of the disk D1(−i2)=D_{1}(-\frac{i}{2})={(u,v) ∣ u2+(v+12)2>1(u,v)\space |\space u^2+(v+\frac{1}{2})^2>1}.

16. y<x−12→f(z)=1z∣w−1−i∣<2y<x-\frac{1}{2} \xrightarrow{f(z)=\frac{1}{z}}|w-1-i|<\sqrt{2}

−vu2+v2<uu2+v2−12⇔u2−2u+v2−2v<0\frac{-v}{u^2+v^2}<\frac{u}{u^2+v^2}-\frac{1}{2} \Leftrightarrow u^2-2u+v^2-2v<0

⇔(u−1)2+(v−1)2<2\Leftrightarrow (u-1)^2+(v-1)^2<2

which is the disk D2(1+i)D_{\sqrt{2}}(1+i).

D2(1+i)=D_{\sqrt{2}}(1+i)={(u,v):∣w−1−i∣<2(u,v): |w-1-i|<\sqrt{2}}

  1. 2x=1−y2→f(z)=1zρ=1+cos⁡ϕ2x=1-y^2 \xrightarrow{f(z)=\frac{1}{z}} \rho =1+\cos{\phi}

    2uu2+v2=1−(−vu2+v2)2,w=u+iv=ρeiϕ≠0,u=ρcos⁡ϕ,v=ρsin⁡ϕ\frac{2u}{u^2+v^2}=1-(\frac{-v}{u^2+v^2})^2, w=u+iv=\rho e^{i\phi} \neq 0, u=\rho \cos{\phi}, v=\rho \sin{\phi}

    ⇔2u(u2+v2)=(u2+v2)2−v2\Leftrightarrow 2u(u^2+v^2)=(u^2+v^2)^2-v^2

    ⇔2ρcos⁡ϕ⋅ρ2=ρ4−ρ2sin⁡2ϕ\Leftrightarrow 2\rho \cos{\phi} \cdot \rho^2=\rho^4-\rho^2 \sin^2{\phi}

    ⇔ρ2−2ρcos⁡ϕ−sin⁡2ϕ=0\Leftrightarrow \rho^2-2\rho \cos{\phi}-\sin^2{\phi}=0

    ρ=2cos⁡ϕ±42=cos⁡ϕ±1\rho =\frac{2\cos{\phi} \pm \sqrt{4}}{2}=\cos{\phi} \pm 1 (取正), ρ>0\rho >0

  1. ∀ ε>0\forall \space \varepsilon >0, ∃ R=2ε+1>0\exist \space R=\frac{2}{\varepsilon}+1>0, ∣z∣>R=2ε+1>0|z|>R=\frac{2}{\varepsilon}+1>0

    ⇒∣z−1∣≥∣z∣−1>R−1=2ε⇔1∣z−1∣<ε2\Rightarrow |z-1| \geq |z|-1>R-1=\frac{2}{\varepsilon} \Leftrightarrow \frac{1}{|z-1|}<\frac{\varepsilon}{2}

    such that ∣z+1z−1−1∣=∣(z+1)−(z−1)z−1∣=2∣z−1∣|\frac{z+1}{z-1}-1|=|\frac{(z+1)-(z-1)}{z-1}|=\frac{2}{|z-1|}

    =2⋅1∣z−1∣<2⋅ε2=ε=2 \cdot \frac{1}{|z-1|}<2 \cdot \frac{\varepsilon}{2}=\varepsilon