2.1 參考解答

  1. (c) f(z)=f(x+iy)=f(1+i)=1+1+i(1−1)=2f(z)=f(x+iy)=f(1+i)=1+1+i(1-1)=2.
  1. (b) f(1+i3)=f(2eiπ3)=(2eiπ3)21−5(2eiπ3)7+9(2eiπ3)4f(1+i\sqrt{3})=f(2e^{i\frac{\pi}{3}})=(2e^{i\frac{\pi}{3}})^{21}-5(2e^{i\frac{\pi}{3}})^7+9(2e^{i\frac{\pi}{3}})^4

    =−2097544−i3923=-2097544-i392\sqrt{3}

  1. (b) z=x+iyz=x+iy

    z‾=x−iy\overline{z}=x-iy

    f(z)=z‾2+(2−3i)zf(z)=\overline{z}^2+(2-3i)z

    =(x−iy)2+(2−3i)(x+iy)=(x-iy)^2+(2-3i)(x+iy)

    =x2−i2xy−y2+2x+i2y−i3x+3y=x^2-i2xy-y^2+2x+i2y-i3x+3y

    =(x2−y2+2x+3y)+i(−2xy+2y−3x)=(x^2-y^2+2x+3y)+i(-2xy+2y-3x)

  1. (a) z=∣z∣eiθ=reiθz=|z|e^{i\theta}=re^{i\theta}

z‾=∣z∣e−iθ=re−iθ\overline{z}=|z|e^{-i\theta}=re^{-i\theta}

f(z)=z5+z‾5f(z)=z^5+\overline{z}^5

=(reiθ)5+(re−iθ)5=(re^{i\theta})^5+(re^{-i\theta})^5

=r5ei5θ+r5e−i5θ=r^5e^{i5\theta}+r^5e^{-i5\theta}

=r5[cos⁡(5θ)+isin⁡(5θ)+cos⁡(−5θ)+isin⁡(−5θ)]=r^5[\cos{(5\theta)}+i\sin{(5\theta)}+\cos{(-5\theta)}+i\sin{(-5\theta)}]

=2r5cos⁡(5θ)=2r^5\cos{(5\theta)}

6. (c) f(1+i3)=12ln⁡(1+3)+itan⁡−131f(1+i\sqrt{3})=\frac{1}{2}\ln{(1+3)}+i\tan^{-1}{\frac{\sqrt{3}}{1}}

=12ln⁡4+itan⁡−13=\frac{1}{2}\ln{4}+i\tan^{-1}{\sqrt{3}}

=ln⁡2+iπ3=\ln{2}+i\frac{\pi}{3}

  1. (a) f(1)=ln⁡1=0f(1)=\ln{1}=0

    (b) f(−2)=ln⁡2+iπf(-2)=\ln{2}+i\pi

    (c) f(1+i)=ln⁡12+iπ4=12ln⁡2+iπ4f(1+i)=\ln{1\sqrt{2}}+i\frac{\pi}{4}=\frac{1}{2}\ln{2}+i\frac{\pi}{4}

    (d) f(−3+i)=ln⁡2+i56πf(-\sqrt{3}+i)=\ln{2}+i\frac{5}{6}\pi

    (e) Yes.

    If f(z1)=f(z2)f(z_1)=f(z_2), then ln⁡r1+iθ1=ln⁡r2+iθ2\ln{r_1}+i\theta_1=\ln{r_2}+i\theta_2.

    ⇒ln⁡r1=ln⁡r2\Rightarrow \ln{r_1}=\ln{r_2} and iθ1=iθ2i\theta_1=i\theta_2

    So r1=r2r_1=r_2 and θ1=θ2\theta_1=\theta_2. (ln⁡\ln is 1-1)

    i.e. z1=z2z_1=z_2.

  1. (a) Let f(z1)=f(z2)f(z_1)=f(z_2)

    g(f(z1))=z1g(f(z_1))=z_1

    g(f(z2))=z2g(f(z_2))=z_2

    ⇒z1=z2\Rightarrow z_1=z_2

    ∴f\therefore f is 1-1.

    (b) Let b∈Bb\in B, g(b)∈A⇒f(g(b))=bg(b)\in A \Rightarrow f(g(b))=b.

    ∴f\therefore f is an onto map.

10. f(z)=(3+4i)z−2+i⇒f−1(w)=w+2−i3+4i=zf(z)=(3+4i)z-2+i \Rightarrow f^{-1}(w)=\frac{w+2-i}{3+4i}=z

(a) ∣z−1∣<1|z-1|<1

⇒∣w+2−i3+4i−1∣<1\Rightarrow |\frac{w+2-i}{3+4i}-1|<1

⇒∣w+2−i−(3+4i)∣<∣3+4i∣=5\Rightarrow |w+2-i-(3+4i)|<|3+4i|=5

⇒∣w−(1+5i)∣<5\Rightarrow |w-(1+5i)|<5

(b) x=t,y=1−2t⇒2x+y−1=0x=t, y=1-2t \Rightarrow 2x+y-1=0

f(x+iy)=(3+4i)(x+iy)−2+i=(3x−4y−2)+i(4x+3y+1)f(x+iy)=(3+4i)(x+iy)-2+i=(3x-4y-2)+i(4x+3y+1)

u=3x−4y−2,v=4x+3y+1⇒{u=11t−6v=−2t+4u=3x-4y-2, v=4x+3y+1 \Rightarrow \begin{cases} u=11t-6 \\ v=-2t+4 \end{cases}

⇒2u+11v=32\Rightarrow 2u+11v=32

(c) Im(z)>1\text{Im}(z)>1

z=w+2−i3+4i⋅3−4i3−4i=125[(3u+4v+2)+i(−4u+3v−11)]z=\frac{w+2-i}{3+4i} \cdot \frac{3-4i}{3-4i}=\frac{1}{25}[(3u+4v+2)+i(-4u+3v-11)]

∴−4u+3v−1125>1⇒4u−3v<36\therefore \frac{-4u+3v-11}{25}>1 \Rightarrow 4u-3v<36.

  1. (a) z1=2→w1=1+i→w2=1z_1=2 \rightarrow w_1=1+i \rightarrow w_2=1

    Let f(z)=Az+Bf(z)=Az+B.

    {f(2)=2A+B=1+if(−3i)=(−3i)A+B=1⇒{A=113(3+2i)B=113(7+9i)\begin{cases} f(2)=2A+B=1+i \\ f(-3i)=(-3i)A+B=1 \end{cases} \Rightarrow \begin{cases} A=\frac{1}{13}(3+2i) \\ B=\frac{1}{13}(7+9i) \end{cases}

    ∴f(z)=113(3+2i)z+113(7+9i)\therefore f(z)=\frac{1}{13}(3+2i)z+\frac{1}{13}(7+9i)

    (b) ∣z∣=1→∣w−3+2i∣=5,f(−i)=3+3i|z|=1 \rightarrow |w-3+2i|=5, f(-i)=3+3i

    Let f(z)=Az+Bf(z)=Az+B.

    {f(0)=A⋅0+B=3−2if(−i)=A⋅(−i)+B=3+3i⇒{A=−5B=3−2i\begin{cases} f(0)=A \cdot 0+B=3-2i \\ f(-i)=A \cdot (-i)+B=3+3i \end{cases} \Rightarrow \begin{cases} A=-5 \\ B=3-2i \end{cases}

    ∴f(z)=−5z+(3−2i)\therefore f(z)=-5z+(3-2i)

    (c) z1=−4+2i→w1=1,z2=−4+7i→w2=0,z3=1+2i→w3=1+iz_1=-4+2i \rightarrow w_1=1, z_2=-4+7i \rightarrow w_2=0, z_3=1+2i \rightarrow w_3=1+i

    Let f(z)=Az+Bf(z)=Az+B.

    {f(−4+2i)=A⋅(−4+2i)+B=1f(−4+7i)=A⋅(−4+7i)+B=0f(1+2i)=A⋅(1+2i)+B=1+i\begin{cases} f(-4+2i)=A \cdot (-4+2i)+B=1 \\ f(-4+7i)=A \cdot (-4+7i)+B=0 \\ f(1+2i)=A \cdot (1+2i)+B=1+i \end{cases}

    ⇒A=i5,B=15(7+4i)\Rightarrow A=\frac{i}{5}, B=\frac{1}{5}(7+4i)

    ∴f(z)=i5z+15(7+4i)\therefore f(z)=\frac{i}{5}z+\frac{1}{5}(7+4i)