2.2 參考解答

  1. w=f(z)=z2w=f(z)=z^2

    Let z=x+iy⇒w=(x+iy)2=x2−y2+i2xy=u+ivz=x+iy \Rightarrow w=(x+iy)^2=x^2-y^2+i2xy=u+iv

    ⇒{u=x2−y2v=2xy\Rightarrow \begin{cases} u=x^2-y^2 \\ v=2xy \end{cases}

    (d) ***

    (f) ***

    ⇒{u=x2−y22xy⇒u=1\Rightarrow \begin{cases} u=x^2-y^2 \\ 2xy \end{cases} \Rightarrow u=1

    A={x+iy ∣ x>0,x2−y2≥1}A=\{x+iy\space|\space x>0, x^2-y^2 \geq 1\}

    f(A)={u+iv ∣ u≥1}f(A)=\{u+iv\space|\space u \geq 1\}

3. (a) 2z2+5iz−2=02z^2+5iz-2=0

⇒(2z+i)(z+2i)=0\Rightarrow (2z+i)(z+2i)=0

⇒z=−i2\Rightarrow z=-\frac{i}{2} or −2i-2i

(b) 3z2−10z+3=03z^2-10z+3=0

⇒(3z−1)(z−3)=0\Rightarrow (3z-1)(z-3)=0

⇒z=13\Rightarrow z=\frac{1}{3} or 33

(c) z2+2z+5=0z^2+2z+5=0

⇒(z+1)2+4=0\Rightarrow (z+1)^2+4=0

⇒(z+1)2=−4\Rightarrow (z+1)^2=-4

⇒z+1=±2i\Rightarrow z+1=\pm 2i

⇒z=−1±2i\Rightarrow z=-1\pm 2i

(d) 2z2+5iz−2=02z^2+5iz-2=0

⇒z=−2±4−84=−1±i2\Rightarrow z=\frac{-2\pm \sqrt{4-8}}{4}=\frac{-1\pm i}{2}

  1. (a) Re(z2)>4\text{Re}(z^2)>4

    z2=(x2−y2)+i2xyz^2=(x^2-y^2)+i2xy

    Re(z2)=x2−y2>4\text{Re}(z^2)=x^2-y^2>4

9. Re(z)>1,w=z2+2z+1\text{Re}(z)>1, w=z^2+2z+1

Let w=f(z)=z2+2z+1w=f(z)=z^2+2z+1, z=x+iyz=x+iy

=(x+iy)2+2(x+iy)+1=(x+iy)^2+2(x+iy)+1

=(x2−y2+2x+1)+i(2xy+2y)=(x^2-y^2+2x+1)+i(2xy+2y)

⇒{u=x2−y2+2x+1=(x+1)2−y2v=2xy+2y=2y(x+1)\Rightarrow \begin{cases} u=x^2-y^2+2x+1=(x+1)^2-y^2 \\ v=2xy+2y=2y(x+1) \end{cases}

Re(z)=1\text{Re}(z)=1, x=1⇒{u=4−y2v=4y⇒u=4−v216x=1 \Rightarrow \begin{cases} u=4-y^2 \\ v=4y \end{cases} \Rightarrow u=4-\frac{v^2}{16}

The region in ww-plane that lies to the right of parabola u=4−v216u=4-\frac{v^2}{16}.

  1. (a) w=z3={reiθ:r>8w=z^3=\{re^{i\theta}: r>8, and 3π4<θ<π}\frac{3\pi}{4}<\theta<\pi\}

    <解> w=z3w=z^3

    ρeiϕ=(reiθ)3=r3ei3θ\rho e^{i\phi}=(re^{i\theta})^3=r^3e^{i3\theta}

    r>2⇒r3>8r>2 \Rightarrow r^3>8

    π4<θ<π3⇒3π4<θ<π\frac{\pi}{4}<\theta<\frac{\pi}{3} \Rightarrow \frac{3\pi}{4}<\theta<\pi

    ∴{w=ρeiϕ:ρ>8 \therefore \{w=\rho e^{i\phi}: \rho>8 and 3π4<ϕ<π}\frac{3\pi}{4}<\phi<\pi\}

    (b) w=z4={reiθ:r>16w=z^4=\{re^{i\theta}: r>16, and π<θ<4π3}\pi<\theta<\frac{4\pi}{3}\}

    (c) w=z6={reiθ:r>64w=z^6=\{re^{i\theta}: r>64, and 3π2<θ<2π}\frac{3\pi}{2}<\theta<2\pi\}

  1. (a) w=z12={reiθ:r>0w=z^\frac{1}{2}=\{re^{i\theta}: r>0, and −π2<θ<π3}-\frac{\pi}{2}<\theta<\frac{\pi}{3}\}

    <解> w=z13⇒w3=zw=z^\frac{1}{3} \Rightarrow w^3=z

    ⇒(ρeiϕ)3=ρ3ei3ϕ=reiθ\Rightarrow (\rho e^{i\phi})^3=\rho^3 e^{i3\phi}=re^{i\theta}

    r>0⇒ρ3>0⇒ρ>0r>0 \Rightarrow \rho^3>0 \Rightarrow \rho >0

    −π<θ<2π3⇒−π<3ϕ<2π3-\pi<\theta<\frac{2\pi}{3} \Rightarrow -\pi<3\phi<\frac{2\pi}{3}

    ⇒−π3<ϕ<2π9\Rightarrow -\frac{\pi}{3}<\phi<\frac{2\pi}{9}

    ∴{w=ρeiϕ:ρ>0\therefore \{w=\rho e^{i\phi}: \rho>0 and −π3<ϕ<2π9}-\frac{\pi}{3}<\phi<\frac{2\pi}{9}\}

    (b) w=z13={reiθ:r>0w=z^\frac{1}{3}=\{re^{i\theta}: r>0, and −1π3<θ<2π9}-\frac{1\pi}{3}<\theta<\frac{2\pi}{9}\}

    (c) w=z14={reiθ:r>0w=z^\frac{1}{4}=\{re^{i\theta}: r>0, and −π4<θ<π6}-\frac{\pi}{4}<\theta<\frac{\pi}{6}\}