5.2 參考解答

  1. (d) ∗Log(z)=ln⁡∣z∣+iArg (z)\ast \text{Log}(z)=\ln{|z|}+i\text{Arg}\space (z), −π<Arg (z)≤π-\pi<\text{Arg}\space (z) \leq \pi

    Log[(1+i)4]=Log(−4)=ln⁡∣−4∣+iArg(−4)=ln⁡4+iπ\text{Log}[(1+i)^4]=\text{Log}(-4)=\ln{|-4|}+i\text{Arg}(-4)=\ln{4}+i\pi

    (h) ∗log⁡(z)=ln⁡∣z∣+iarg⁡(z)\ast \log(z)=\ln{|z|}+i\arg(z)

    log⁡(−3−i)=ln⁡∣−3−i∣+iarg⁡(−3−i)\log{(-\sqrt{3}-i})=\ln{|-\sqrt{3}-i|}+i\arg{(-\sqrt{3}-i)}

    =ln⁡2+i(−5π6+2nπ)=\ln{2}+i(-\frac{5\pi}{6}+2n\pi), n∈Zn \in \mathbb{Z}

    =ln⁡2+i(−56+2n)π=\ln{2}+i(-\frac{5}{6}+2n)\pi, n∈Zn \in \mathbb{Z}

3. (b) Log(z−1)=ln⁡∣z−1∣+iArg(z−1)=iπ2\text{Log}(z-1)=\ln{|z-1|}+i\text{Arg}(z-1)=i\frac{\pi}{2}

{ln⁡∣z−1∣=0⇒∣z−1∣=1Arg (z)=π2\begin{cases} \ln{|z-1|}=0 \Rightarrow |z-1|=1 \\ \text{Arg}\space (z)=\frac{\pi}{2} \end{cases}

Let z=x+iyz=x+iy

{∣(x−1)+iy∣=1⇒(x−1)2+y2=1⇒y2=1⇒y±1(負不合)Arg [(x−1)+iy]=π2⇒x−1=0⇒x=1 \begin{cases} |(x-1)+iy|=1 \Rightarrow (x-1)^2+y^2=1 \Rightarrow y^2=1 \Rightarrow y\pm 1 \text{(負不合)}\\ \text{Arg}\space [(x-1)+iy]=\frac{\pi}{2} \Rightarrow x-1=0 \Rightarrow x=1 \end{cases}

∴z=1+i\therefore z=1+i

(d) ez+1=i⇒log⁡(i)=z+1e^{z+1}=i \Rightarrow \log{(i)}=z+1

Let z=x+iyz=x+iy

log⁡(i)=(x+1)+iy\log{(i)}=(x+1)+iy

⇒ln⁡∣i∣+iarg⁡(i)=i(π2+2nπ)\Rightarrow \ln{|i|}+i\arg{(i)}=i(\frac{\pi}{2}+2n\pi), n∈Zn \in \mathbb{Z}

{x+1=0⇒x=−1y=π2+2nπ,n∈Z\begin{cases} x+1=0 \Rightarrow x=-1 \\ y=\frac{\pi}{2}+2n\pi, n \in \mathbb{Z} \end{cases}

∴z=−1+i(π2+2nπ),n∈Z\therefore z=-1+i(\frac{\pi}{2}+2n\pi), n \in \mathbb{Z}

  1. ∗log⁡α(z)=ln⁡r+iθ\ast \log_\alpha {(z)}=\ln{r}+i\theta, α<θ≤α+2π\alpha<\theta \leq \alpha+2\pi

    (b) z0=−1+i3z_0=-1+i\sqrt{3}

    ⇒r=2\Rightarrow r=2, θ=2π3\theta=\frac{2\pi}{3}

    ∴α=2π3+2kπ\therefore \alpha=\frac{2\pi}{3}+2k\pi, k∈Zk \in \mathbb{Z}

    (d) z0=−iz_0=-i

    ⇒r=1\Rightarrow r=1, θ=−π2\theta=-\frac{\pi}{2}

    ∴α=−π2+2kπ\therefore \alpha=-\frac{\pi}{2}+2k\pi, k∈Zk \in \mathbb{Z}

  1. f(z)f(z) is analytic everywhere except −1-1, −2-2.

    ( ∵z2+3z+2=(z+1)(z+2)\because z^2+3z+2=(z+1)(z+2) )

    The composite function Log(g(z))\text{Log}(g(z)), where g(z)=z+5=x+5g(z)=z+5=x+5,

    is analytic except at those points where g(z)=0g(z)=0,

    or where zz lies on the negative real axis, {(x,y):x≤−5,y=0}\{(x,y): x \leq -5, y=0\}.

9. By log⁡α(z)=ln⁡∣z∣+iθ\log_\alpha {(z)}=\ln{|z|}+i\theta, α<θ≤α+2π\alpha<\theta \leq \alpha+2\pi

Let α=6π\alpha=6\pi

f(−5)=log⁡6π(−1)=ln⁡1+i(7π)=7πif(-5)=\log_{6\pi}{(-1)}=\ln{1}+i(7\pi)=7\pi i

  1. (b) Let C={z=2eiθ:−π2≤θ≤π2}C=\{z=2e^{i\theta}: -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\}

    D={(ln⁡2,v):−π2≤v≤π2}D=\{(\ln{2},v): -\frac{\pi}{2} \leq v \leq \frac{\pi}{2}\}

    ∗ Log(z)=ln⁡∣z∣+iArg (z)\ast\space \text{Log}(z)=\ln{|z|}+i\text{Arg}\space (z)

    ∵z=2eiθ\because z=2e^{i\theta}, −π2≤θ≤π2⇒∣z∣=2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\Rightarrow |z|=2, Arg (z)=θ\text{Arg}\space (z)=\theta

    ∴Log(z)=Log(2eiθ)=ln⁡2+iθ\therefore \text{Log}(z)=\text{Log}(2e^{i\theta})=\ln{2}+i\theta, −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

    1。1^。 1-1

    Assume Log(z1)=Log(z2)\text{Log}(z_1)=\text{Log}(z_2), for z1z_1, z2∈Cz_2 \in \mathbb{C}

    i.e. z1=2eiθ1z_1=2e^{i\theta_1}, z2=2eiθ2z_2=2e^{i\theta_2}, −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

    ⇒ln⁡2+iθ1=ln⁡2+iθ2\Rightarrow \ln{2}+i\theta_1=\ln{2}+i\theta_2

    ⇒θ1=θ2\Rightarrow \theta_1=\theta_2

    ⇒z1=z2\Rightarrow z_1=z_2

    ∴Log(z)\therefore \text{Log}(z) is 1-1 from CC to DD.

    2。2^。 onto

    ∀ w∈D⇒∃ v \forall \space w \in D \Rightarrow \exist \space v such that w=ln⁡2+ivw=\ln{2}+iv, −π2≤θ≤π2-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

    ⇒∀ z=2eiv\Rightarrow \forall \space z=2e^{iv}, Log(z)=ln⁡∣z∣+iArg (z)\text{Log}(z)=\ln{|z|}+i\text{Arg}\space (z)

    =ln⁡2+iv=w=\ln{2}+iv=w

    ∴∀ w∈D\therefore \forall \space w \in D, ∃ z∈C\exist \space z \in C such that Log(z)=w\text{Log}(z)=w.

    ∴Log(z)\therefore \text{Log}(z) is onto from CC to DD.