5.1 參考解答

  1. ez=ex+iy=ex(cos⁡y+isin⁡y)e^z=e^{x+iy}=e^x(\cos{y}+i\sin{y})

    (c) z=−4+5iz=-4+5i (i.e. z=−4+5iz=-4+5i)

    e−4+5i=e−4(cos⁡5+isin⁡5)e^{-4+5i}=e^{-4}(\cos{5}+i\sin{5})

    ≈0.00519147−i0.0175623\approx 0.00519147-i0.0175623

    (e) z=1+i5π4z=1+i\frac{5\pi}{4}

    e1+i5π4=e(cos⁡5π4+isin⁡5π4)e^{1+i\frac{5\pi}{4}}=e(\cos{\frac{5\pi}{4}}+i\sin{\frac{5\pi}{4}})

    =e(−22−i22) =e(-\frac{\sqrt{2}}{2}-i\frac{\sqrt{2}}{2})

  1. ez=ex+iy=ex(cos⁡y+isin⁡y)e^z=e^{x+iy}=e^x(\cos{y}+i\sin{y})

    (a) e^z=-4

    {excos⁡y=−4exsin⁡y=0⇒{ex=4, x=ln⁡4sin⁡y=0, y=(2n+1)π, n∈Z\begin{cases} e^x\cos{y}=-4 \\ e^x\sin{y}=0 \end{cases} \Rightarrow \begin{cases} e^x=4,\space x=\ln{4} \\ \sin{y}=0,\space y=(2n+1)\pi,\space n \in \mathbb{Z} \end{cases}

    ∴z=ln⁡4+i(2n+1)π\therefore z=\ln{4}+i(2n+1)\pi, n∈Zn \in \mathbb{Z}

    (c) ez=3−ie^z=\sqrt{3}-i

    {excos⁡y=3−−(a)exsin⁡y=−1−−(b)\begin{cases} e^x\cos{y}=\sqrt{3}--\text{(a)} \\ e^x\sin{y}=-1 --\text{(b)}\end{cases}

    (a)2+(b)2⇒e2x=4⇒ex=2\text{(a)}^2+\text{(b)}^2 \Rightarrow e^{2x}=4 \Rightarrow e^x=2, x=ln⁡2x=\ln{2}

    ∴cos⁡y=32,sin⁡y=−12\therefore \cos{y}=\frac{\sqrt{3}}{2}, \sin{y}=-\frac{1}{2}, y=−π6+2nπy=-\frac{\pi}{6}+2n\pi, n∈Zn \in \mathbb{Z}

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

  1. Prove ∣ez2∣≤e∣z∣2|e^{z^2}| \leq e^{|z|^2} for all z∈Cz \in \mathbb{C}.

    Let z=x+iyz=x+iy

    ∣ez2∣=∣ex2−y2+i2xy∣=∣ex2−y2∣∣ei2xy∣=∣ex2−y2∣≤∣x2+y2∣=e∣z∣2|e^{z^2}|=|e^{x^2-y^2+i2xy}|=|e^{x^2-y^2}||e^{i2xy}|=|e^{x^2-y^2}| \leq |x^2+y^2|=e^{|z|^2}

    ∴∣ez2∣≤e∣z∣2\therefore |e^{z^2}| \leq e^{|z|^2}

    “==” y=0y=0, 則等式成立

8. 1。1^。 ez2=ex2−y2+i2xy=ex2−y2(cos⁡2xy+isin⁡2xy)e^{z^2}=e^{x^2-y^2+i2xy}=e^{x^2-y^2}(\cos{2xy}+i\sin{2xy})

∴u(x,y)=ex2−y2cos⁡2xy\therefore u(x,y)=e^{x^2-y^2}\cos{2xy}, v(x,y)=ex2−y2sin⁡2xyv(x,y)=e^{x^2-y^2}\sin{2xy}

2。2^。 e1z=e1x+iy=ex−iyx2+y2=exx2+y2⋅ei(−yx2+y2)e^{\frac{1}{z}}=e^{\frac{1}{x+iy}}=e^{\frac{x-iy}{x^2+y^2}}=e^{\frac{x}{x^2+y^2}} \cdot e^{i(\frac{-y}{x^2+y^2})}

∴u(x,y)=exx2+y2cos⁡(−yx2+y2)=exx2+y2cos⁡(yx2+y2)\therefore u(x,y)=e^{\frac{x}{x^2+y^2}} \cos{(\frac{-y}{x^2+y^2})}=e^{\frac{x}{x^2+y^2}} \cos{(\frac{y}{x^2+y^2})}

v(x,y)=exx2+y2sin⁡(−yx2+y2)=−exx2+y2sin⁡(yx2+y2)v(x,y)=e^{\frac{x}{x^2+y^2}} \sin{(\frac{-y}{x^2+y^2})}=-e^{\frac{x}{x^2+y^2}} \sin{(\frac{y}{x^2+y^2})}

  1. (b) by L’Hoˆ\^{\text{o}}spital: lim⁡z→iπez+1z−iπ=lim⁡z→iπez1=eiπ=−1\lim_{z \to i\pi} \frac{e^z+1}{z-i\pi}=\lim_{z \to i\pi} \frac{e^z}{1}=e^{i\pi}=-1

13. (b) ddz(z4ez3)=4z3ez3+z4⋅3z2ez3=ez3(4z3+3z6)\frac{d}{dz} (z^4 e^{z^3})=4z^3e^{z^3}+z^4\cdot 3z^2e^{z^3}=e^{z^3}(4z^3+3z^6)

(d) ddz(e1z)=−1z2e1z\frac{d}{dz} (e^{\frac{1}{z}})=-\frac{1}{z^2}e^{\frac{1}{z}}

  1. (a) n∈N,(ez)n=enzn \in \mathbb{N}, (e^z)^n=e^{nz}

    Let z=x+iyz=x+iy

    (ez)n=(ex+iy)n=(ex(cos⁡y+isin⁡y))n(e^z)^n=(e^{x+iy})^n=(e^x(\cos{y}+i\sin{y}))^n

    =enx(cos⁡y+isin⁡y)n=e^{nx}(\cos{y}+i\sin{y})^n

    =enx(cos⁡ny+isin⁡ny)=e^{nx}(\cos{ny}+i\sin{ny})

    =enx⋅einy=e^{nx}\cdot e^{iny}

    =enx+iny=e^{nx+iny}

    =en(x+iy)=e^{n(x+iy)}

    =enz=e^{nz}

  1. Σn=0∞einz\Sigma_{n=0}^{\infty} e^{inz} converges for Im(z)>0\text{Im}(z)>0

    Σn=0∞einz=Σn=0∞(eiz)n\Sigma_{n=0}^{\infty} e^{inz}=\Sigma_{n=0}^{\infty} (e^{iz})^n if converges ∣eiz∣<1|e^{iz}|<1

    Let z=x+iyz=x+iy, Im(z)>0⇒y>0\text{Im}(z)>0 \Rightarrow y>0, ey>1⇒e−y<1e^y>1 \Rightarrow e^{-y}<1

    ∣e−y∣∣eix∣=∣e−y+ix∣=∣ei(x+iy)∣=∣eiz∣<1|e^{-y}||e^{ix}|=|e^{-y+ix}|=|e^{i(x+iy)}|=|e^{iz}|<1

18. (a) f(z)=ezf(z)=e^z

{(x,y):x=t,y=2π+t}\{(x,y): x=t, y=2\pi+t\}, t∈Rt \in \mathbb{R}

Let z=x+iyz=x+iy

f(z)=ex+iy=ex(cos⁡y+isin⁡y)f(z)=e^{x+iy}=e^x(\cos{y}+i\sin{y})

=et(cos⁡2π+t+isin⁡2π+t)=e^t(\cos{2\pi+t}+i\sin{2\pi+t})

=et(cos⁡t+isin⁡t)=u+iv=e^t(\cos{t}+i\sin{t})=u+iv

u=etcos⁡t,v=etsin⁡tu=e^t\cos{t}, v=e^t\sin{t}

(b) {(x,y):x=2,y=t}\{(x,y): x=2, y=t\}

f(z)=ex+iy=ex(cos⁡y+isin⁡y)=e2(cos⁡t+isin⁡t)f(z)=e^{x+iy}=e^x(\cos{y}+i\sin{y})=e^2(\cos{t}+i\sin{t}), π6<t<7π6\frac{\pi}{6}<t<\frac{7\pi}{6}

(c) {(x,y):x>0,y=π3}\{(x,y): x>0, y=\frac{\pi}{3}\}

f(z)=ex+iy=ex(cos⁡y+isin⁡y)=ex(cos⁡π3+isin⁡π3) f(z)=e^{x+iy}=e^x(\cos{y}+i\sin{y})=e^x(\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}})