3.1 參考解答

  1. (a) f(z)=5z3−4z2+7z−8f(z)=5z^3-4z^2+7z-8

    f’(z)=15z2−8z+7f’(z)=15z^2-8z+7

    (b) g(z)=(z2−iz+9)5g(z)=(z^2-iz+9)^5

    g’(z)=5(z2−iz+9)4(2z−i)g’(z)=5(z^2-iz+9)^4(2z-i)

    (c) h(z)=2z+1z+2h(z)=\frac{2z+1}{z+2} for z≠−2z \neq -2

    h’(z)=2(z+2)−(2z+1)⋅1(z+2)2=3(z+2)2h’(z)=\frac{2(z+2)-(2z+1)\cdot 1}{(z+2)^2}=\frac{3}{(z+2)^2} for z≠−2z \neq -2

    (d) F(z)=[z2+(1−3i)z+1][z4+3z2+5i]F(z)=[z^2+(1-3i)z+1][z^4+3z^2+5i]

    F’(z)=[2z+(1−3i)][z4+3z2+5i]+[z2+(1−3i)z+1][4z3+6z]F’(z)=[2z+(1-3i)][z^4+3z^2+5i]+[z^2+(1-3i)z+1][4z^3+6z]

  1. (a) f(z)=Re(zf(z)=\text{Re}(z), let z=x+iyz=x+iy

    1。1^。 lim⁡z→z0y=0Re(z)−Re(z0)z−z0=lim⁡(x,y)→(x0,y0)y=0x−x0(x+iy)−(x0+iy0)=lim⁡x→x0x−x0x−x0=1\lim_{z \to z_0 \atop y=0} \frac{\text{Re}(z)-\text{Re}(z_0)}{z-z_0}=\lim_{(x,y) \to (x_0,y_0) \atop y=0} \frac{x-x_0}{(x+iy)-(x_0+iy_0)}=\lim_{x \to x_0} \frac{x-x_0}{x-x_0}=1

    2。2^。 lim⁡z→z0x=0Re(z)−Re(z0)z−z0=lim⁡(x,y)→(x0,y0)x=0x−x0(x+iy)−(x0+iy0)=lim⁡x→x0x−x0i(y−y0)=0\lim_{z \to z_0 \atop x=0} \frac{\text{Re}(z)-\text{Re}(z_0)}{z-z_0}=\lim_{(x,y) \to (x_0,y_0) \atop x=0} \frac{x-x_0}{(x+iy)-(x_0+iy_0)}=\lim_{x \to x_0} \frac{x-x_0}{i(y-y_0)}=0

    By 1。≠2。1^。\neq 2^。, Re(z)\text{Re}(z) is differentiable nowhere.

    (b) f(z)=Im(z)f(z)=\text{Im}(z), let z=x+iyz=x+iy

    1。1^。 lim⁡z→z0x=0Im(z)−Im(z0)z−z0=lim⁡(x,y)→(x0,y0)x=0y−y0(x+iy)−(x0+iy0)=lim⁡y→y0y−y0i(y−y0)=−i\lim_{z \to z_0 \atop x=0} \frac{\text{Im}(z)-\text{Im}(z_0)}{z-z_0}=\lim_{(x,y) \to (x_0,y_0) \atop x=0} \frac{y-y_0}{(x+iy)-(x_0+iy_0)}=\lim_{y \to y_0} \frac{y-y_0}{i(y-y_0)}=-i

    2。2^。 lim⁡z→z0y=0Re(z)−Re(z0)z−z0=lim⁡(x,y)→(x0,y0)y=0y−y0(x+iy)−(x0+iy0)=lim⁡y→y0y−y0x−x0=0\lim_{z \to z_0 \atop y=0} \frac{\text{Re}(z)-\text{Re}(z_0)}{z-z_0}=\lim_{(x,y) \to (x_0,y_0) \atop y=0} \frac{y-y_0}{(x+iy)-(x_0+iy_0)}=\lim_{y \to y_0} \frac{y-y_0}{x-x_0}=0

    By 1。≠2。1^。\neq 2^。, Im(z)\text{Im}(z) is differentiable nowhere.

  1. (a)(b)(e)(f), (c) provided that g(z)≠0g(z) \neq 0, ∀ z∈C\forall \space z \in \mathbb{C}

6. P(z)=(z−z1)(z−z2)P(z)=(z-z_1)(z-z_2)

P’(z)=(z−z2)+(z−z1)P’(z)=(z-z_2)+(z-z_1)

∴P’(z)P(z)=(z−z2)+(z−z1)(z−z1)(z−z2)=1z−z1+1z−z2\therefore \frac{P’(z)}{P(z)}=\frac{(z-z_2)+(z-z_1)}{(z-z_1)(z-z_2)}=\frac{1}{z-z_1}+\frac{1}{z-z_2}

  1. (a) lim⁡z→iz4−1z−i=lim⁡z→4z31=4(i)3=−4i\lim_{z \to i} \frac{z^4-1}{z-i}=\lim_{z \to } \frac{4z^3}{1}=4(i)^3=-4i

    (b) lim⁡z→1+iz2−iz−2−iz2−2z+2=lim⁡z→1+i2z−i2z−2=2+i2i=12−i\lim_{z \to 1+i} \frac{z^2-iz-2-i}{z^2-2z+2}=\lim_{z \to 1+i} \frac{2z-i}{2z-2}=\frac{2+i}{2i}=\frac{1}{2}-i

    (c) lim⁡z→−iz6+1z2+1=lim⁡z→−i6z52z=−6i−2i=3\lim_{z \to -i} \frac{z^6+1}{z^2+1}=\lim_{z \to -i} \frac{6z^5}{2z}=\frac{-6i}{-2i}=3

    (d) lim⁡z→1+iz4+4z2−2z+2=lim⁡z→1+i4z32z−2=lim⁡z→1+i2z3z−1=4i(1+i)i=4+4i\lim_{z \to 1+i} \frac{z^4+4}{z^2-2z+2}=\lim_{z \to 1+i} \frac{4z^3}{2z-2}=\lim_{z \to 1+i} \frac{2z^3}{z-1}=\frac{4i(1+i)}{i}=4+4i

    (g) lim⁡z→1+i3z6−64z3+8=lim⁡z→1+i36z53z2=2(1+i3)3=−16\lim_{z \to 1+i\sqrt{3}} \frac{z^6-64}{z^3+8}=\lim_{z \to 1+i\sqrt{3}} \frac{6z^5}{3z^2}=2(1+i\sqrt{3})^3=-16

    (f) lim⁡z→−1+i3z9−512z3−8=lim⁡z→−1+i39z83z2=3(−1+i3)6=192\lim_{z \to -1+i\sqrt{3}} \frac{z^9-512}{z^3-8}=\lim_{z \to -1+i\sqrt{3}} \frac{9z^8}{3z^2}=3(-1+i\sqrt{3})^6=192