2.3 參考解答

  1. (c) lim⁡z→iz4−1z−i=lim⁡z→i(z2−1)(z−i)(z+i)z−i=(i2−1)(i+i)=−2⋅2i=−4i\lim_{z \to i} \frac{z^4-1}{z-i}=\lim_{z \to i} \frac{(z^2-1)(z-i)(z+i)}{z-i}=(i^2-1)(i+i)=-2 \cdot 2i=-4i

    (d) lim⁡z→1+iz2+z−2+iz2−2z+1=−1+4i−1=1−4i\lim_{z \to 1+i} \frac{z^2+z-2+i}{z^2-2z+1}=\frac{-1+4i}{-1}=1-4i

    分子：z2+z−2+i=(1+i)2+(1+i)−2+iz^2+z-2+i=(1+i)^2+(1+i)-2+i

    =1+2i−1+1+i−2+i=−1+4i =1+2i-1+1+i-2+i=-1+4i

    分母：z2−2z+1=(z−1)2=(1+i−1)2=i2=−1z^2-2z+1=(z-1)^2=(1+i-1)^2=i^2=-1

    (e) lim⁡z→1+iz2+z−1−3iz2−2z+2=lim⁡z→1+i[z−(−2−i)][z−(1+i)][z−(1+i)][z−(1−i)]=(1+i)−(−2−i)(1+i)−(1−i)=3+2i2i=1−32i\lim_{z \to 1+i} \frac{z^2+z-1-3i}{z^2-2z+2}=\lim_{z \to 1+i} \frac{[z-(-2-i)][z-(1+i)]}{[z-(1+i)][z-(1-i)]}=\frac{(1+i)-(-2-i)}{(1+i)-(1-i)}=\frac{3+2i}{2i}=1-\frac{3}{2}i

    分母：z2−2z+2=z2−2z+1+2−1=(z−1)2−i2z^2-2z+2=z^2-2z+1+2-1=(z-1)^2-i^2

    =[z−(1+i)][z−(1−i)]=[z-(1+i)][z-(1-i)]

    分子：z2+z−1−3i=[z−(a+bi)][z−(1+i)]z^2+z-1-3i=[z-(a+bi)][z-(1+i)]

    ⇒a=−2\Rightarrow a=-2, b=−1b=-1

  1. (d) Let f(z)=z4+1z2+2z+2=z4+1(z+1)2+1f(z)=\frac{z^4+1}{z^2+2z+2}=\frac{z^4+1}{(z+1)^2+1}.

Hence, f(z)f(z) is continuous everywhere.

(e) x+iyx−1\frac{x+iy}{x-1}, continuous everywhere except at {(x,y) ∣ x=1}\{(x,y)\space|\space x=1\}.

(f) x+iy∣z∣−1\frac{x+iy}{|z|-1}, continuous everywhere except at {(x,y) ∣ x2+y2=1}\{(x,y)\space|\space x^2+y^2=1\}.

  1. f(z)={zRe(z)∣z∣,z≠0,0,z=0.f(z)=\begin{cases} \frac{z\text{Re}(z)}{|z|}, z \neq 0, \\ 0, z=0. \end{cases} Show that lim⁡z→0zRe(z)∣z∣=0\lim_{z \to 0} \frac{z\text{Re}(z)}{|z|}=0.

    1。1^。 ∀ ε>0\forall \space \varepsilon>0, ∃ δ≡ε>0\exist \space\delta \equiv \varepsilon >0 such that 0<∣z−0∣=∣z∣<δ0<|z-0|=|z|<\delta

    ⇒∣zRe(z)∣z∣−0∣=∣Re(z)∣≤∣z∣<δ=ε\Rightarrow |\frac{z\text{Re}(z)}{|z|}-0|=|\text{Re}(z)| \leq |z| < \delta =\varepsilon

    ∴lim⁡z→0zRe(z)∣z∣=0\therefore \lim_{z \to 0} \frac{z\text{Re}(z)}{|z|}=0

    2。2^。 zz, Re(z)\text{Re}(z), ∣z∣|z| are continuous function on C\mathbb{C}.

    ⇒zRe(z)∣z∣\Rightarrow \frac{z\text{Re}(z)}{|z|} is continuous for any z≠0z \neq 0.

    By 1。1^。, 2。2^。, lim⁡z→0zRe(z)∣z∣=0=f(0)\lim_{z \to 0} \frac{z\text{Re}(z)}{|z|}=0=f(0), f(z)f(z) is continuous on C\mathbb{C}.

  1. (a) lim⁡z→0z2∣z∣2=lim⁡(x,y)→(0,0)along y=xx2−y2+i2xyx2+y2=lim⁡x→0x2−x2+i2x⋅xx2+x2=lim⁡x→02x2i2x2=i\lim_{z \to 0} \frac{z^2}{|z|^2}=\lim_{(x,y) \to (0,0) \atop \text{along}\space y=x } \frac{x^2-y^2+i2xy}{x^2+y^2}=\lim_{x \to 0} \frac{x^2-x^2+i2x\cdot x}{x^2+x^2}=\lim_{x \to 0} \frac{2x^2i}{2x^2}=i

    (b) lim⁡z→0z2∣z∣2=lim⁡(x,y)→(0,0)along y=2xx2−y2+i2xyx2+y2=lim⁡x→0x2−(2x)2+i2x⋅(2x)x2+(2x)2\lim_{z \to 0} \frac{z^2}{|z|^2}=\lim_{(x,y) \to (0,0) \atop \text{along}\space y=2x } \frac{x^2-y^2+i2xy}{x^2+y^2}=\lim_{x \to 0} \frac{x^2-(2x)^2+i2x\cdot (2x)}{x^2+(2x)^2}

    =lim⁡x→0−3x2+4x2i5x2=−3+4i5=15(−3+4i)=\lim_{x \to 0} \frac{-3x^2+4x^2i}{5x^2}=\frac{-3+4i}{5}=\frac{1}{5}(-3+4i)

    (c) lim⁡z→0z2∣z∣2=lim⁡(x,y)→(0,0)along y=x2x2−y2+i2xyx2+y2=lim⁡x→0x2−(x2)2+i2x⋅x2x2+(x2)2\lim_{z \to 0} \frac{z^2}{|z|^2}=\lim_{(x,y) \to (0,0) \atop \text{along}\space y=x^2 } \frac{x^2-y^2+i2xy}{x^2+y^2}=\lim_{x \to 0} \frac{x^2-(x^2)^2+i2x\cdot x^2}{x^2+(x^2)^2}

    =lim⁡x→0x2−x4+2x3ix2+x4=lim⁡x→01−x2+2xi1+x2=1=\lim_{x \to 0} \frac{x^2-x^4+2x^3i}{x^2+x^4}=\lim_{x \to 0}\frac{1-x^2+2xi}{1+x^2}=1

    (d) No! By (a)(b)(c).

10. ∵−π<Arg z≤π\because -\pi<\text{Arg}\space z \leq \pi

1。1^。 θ=Arg z≤π\theta=\text{Arg}\space z \leq \pi, lim⁡z→−4Arg z=π\lim_{z \to -4} \text{Arg}\space z=\pi

2。 2^。 −π<θ≤Arg z-\pi <\theta \leq \text{Arg}\space z, lim⁡z→−4Arg z=−π\lim_{z \to -4} \text{Arg}\space z=-\pi

By 1。1^。, 2。2^。, lim⁡z→−4Arg z\lim_{z \to -4} \text{Arg}\space z doesn’t exist.

  1. f(z)={Re(z)∣z∣,z≠01,z=0f(z)=\begin{cases} \frac{\text{Re}(z)}{|z|}, z \neq 0 \\ 1, z=0 \end{cases}

    1。1^。 lim⁡z→0Re(z)∣z∣=lim⁡(x,y)→(0,0)along y=xxx2+y2=lim⁡x→0xx2+x2=12\lim_{z \to 0} \frac{\text{Re}(z)}{|z|}=\lim_{(x,y) \to (0,0) \atop \text{along}\space y=x } \frac{x}{\sqrt{x^2+y^2}}=\lim_{x \to 0} \frac{x}{\sqrt{x^2+x^2}}=\frac{1}{\sqrt{2}}

    2。 2^。 lim⁡z→0Re(z)∣z∣=lim⁡(x,y)→(0,0)along y=2xxx2+y2=lim⁡x→0xx2+(2x)2=15\lim_{z \to 0} \frac{\text{Re}(z)}{|z|}=\lim_{(x,y) \to (0,0) \atop \text{along}\space y=2x } \frac{x}{\sqrt{x^2+y^2}}=\lim_{x \to 0} \frac{x}{\sqrt{x^2+(2x)^2}}=\frac{1}{\sqrt{5}}

    By 1。1^。, 2。2^。, lim⁡z→0Re(z)∣z∣2\lim_{z \to 0} \frac{\text{Re}(z)}{|z|^2} doesn’t exist, f(z)f(z) is not continuous on z=0z=0.