5.4 參考解答

  1. ∗cos⁡z=Σn=0∞(−1)nz2n(2n)!\ast \cos{z}=\Sigma_{n=0}^{\infty} (-1)^n\frac{z^{2n}}{(2n)!}, sin⁡z=Σn=0∞(−1)nz2n+1(2n+1)!\sin{z}=\Sigma_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}}{(2n+1)!}

    ddzcos⁡z=ddz[Σn=0∞(−1)nz2n(2n)!]\frac{d}{dz} \cos{z}=\frac{d}{dz} [\Sigma_{n=0}^{\infty} (-1)^n\frac{z^{2n}}{(2n)!}]

    =Σn=0∞[ddz(−1)nz2n(2n)!]=\Sigma_{n=0}^{\infty} [\frac{d}{dz} (-1)^n\frac{z^{2n}}{(2n)!}]

    =Σn=1∞(−1)n(2n)!⋅2n⋅z2n−1=\Sigma_{n=1}^{\infty} \frac{(-1)^n}{(2n)!} \cdot 2n \cdot z^{2n-1}

    =Σn=1∞(−1)nz2n−1(2n−1)!=\Sigma_{n=1}^{\infty} (-1)^n\frac{z^{2n-1}}{(2n-1)!}

    =Σn=0∞(−1)n+1z2n+1(2n+1)!=\Sigma_{n=0}^{\infty} (-1)^{n+1}\frac{z^{2n+1}}{(2n+1)!}

    =−Σn=0∞(−1)nz2n+1(2n+1)!=-\Sigma_{n=0}^{\infty} (-1)^n\frac{z^{2n+1}}{(2n+1)!}

    =−sin⁡z=-\sin{z}

5. ∗sin⁡z=sin⁡xcosh⁡y+icos⁡xsinh⁡y\ast \sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}

sinh⁡z=sinh⁡xcos⁡y+icosh⁡xsin⁡y\sinh{z}=\sinh{x}\cos{y}+i\cosh{x}\sin{y}

(e) Let z=x+iyz=x+iy, iz=−y+ixiz=-y+ix

sin⁡iz=sin⁡(−y)cosh⁡x+icos⁡(−y)sinh⁡x\sin{iz}=\sin{(-y)}\cosh{x}+i\cos{(-y)}\sinh{x}

=−sin⁡ycosh⁡x+icos⁡ysinh⁡x=-\sin{y}\cosh{x}+i\cos{y}\sinh{x}

=i(sinh⁡xcos⁡y+icosh⁡xsin⁡y)=i(\sinh{x}\cos{y}+i\cosh{x}\sin{y})

=isinh⁡z=i\sinh{z}

  1. (b) ∗sin⁡z=sin⁡xcosh⁡y+icos⁡xsinh⁡y\ast \sin{z}=\sin{x}\cosh{y}+i\cos{x}\sinh{y}

    sin⁡π+4i4=sin⁡(π4+i) \sin{\frac{\pi+4i}{4}}=\sin{(\frac{\pi}{4}+i)}

    =sin⁡π4cosh⁡1+icos⁡π4sinh⁡1=\sin{\frac{\pi}{4}}\cosh{1}+i\cos{\frac{\pi}{4}}\sinh{1}

    =22cosh⁡1+i22sinh⁡1=\frac{\sqrt{2}}{2}\cosh{1}+i\frac{\sqrt{2}}{2}\sinh{1}

    =22(cosh⁡1+isinh⁡1)=\frac{\sqrt{2}}{2}(\cosh{1}+i\sinh{1})

    (e) ∗tan⁡z=sin⁡2xcos⁡2x+cosh⁡2y+isinh⁡2ycos⁡2x+cosh⁡2y\ast \tan{z}=\frac{\sin{2x}}{\cos{2x}+\cosh{2y}}+i\frac{\sinh{2y}}{\cos{2x}+\cosh{2y}}

    tan⁡π+2i4=tan⁡(π4+12i)\tan{\frac{\pi+2i}{4}}=\tan{(\frac{\pi}{4}+\frac{1}{2}i)}

    =sin⁡π2cos⁡π2+cosh⁡1+isinh⁡1cos⁡π2+cosh⁡1=\frac{\sin{\frac{\pi}{2}}}{\cos{\frac{\pi}{2}}+\cosh{1}}+i\frac{\sinh{1}}{\cos{\frac{\pi}{2}}+\cosh{1}}

    =1cosh⁡1+isinh⁡1cosh⁡1=\frac{1}{\cosh{1}}+i\frac{\sinh{1}}{\cosh{1}}

    (g) ∗sinh⁡z=sinh⁡xcos⁡y+icosh⁡xsin⁡y\ast \sinh{z}=\sinh{x}\cos{y}+i\cosh{x}\sin{y}

    sinh⁡(1+iπ)=sinh⁡1cos⁡π+icosh⁡1sin⁡π\sinh{(1+i\pi)}=\sinh{1}\cos{\pi}+i\cosh{1}\sin{\pi}

    =−sinh⁡1=-\sinh{1}

10. (e) ∗cosh⁡z=ez+e−z2\ast \cosh{z}=\frac{e^z+e^{-z}}{2}

cosh⁡z=ez+e−z2\cosh{z}=\frac{e^z+e^{-z}}{2}

⇒2cosh⁡z=ez+e−z=2\Rightarrow 2\cosh{z}=e^z+e^{-z}=2

⇒e2z−2ez+1=0\Rightarrow e^{2z}-2e^z+1=0

⇒(ez−1)2=0\Rightarrow (e^z-1)^2=0

⇒ez=1\Rightarrow e^z=1

z=log⁡(1)=ln⁡∣1∣+iarg⁡1=i(2nπ)z=\log{(1)}=\ln{|1|}+i\arg{1}=i(2n\pi), n∈Zn \in \mathbb{Z}

  1. ∗ ϕxx+ϕyy=0\ast\space \phi_{xx}+\phi_{yy}=0

    (b) hx=−sin⁡xsinh⁡yh_x=-\sin{x}\sinh{y}, hy=cos⁡xcosh⁡yh_y=\cos{x}\cosh{y}

    hxx=−cos⁡xsinh⁡yh_{xx}=-\cos{x}\sinh{y}, hyy=cos⁡xsinh⁡yh_{yy}=\cos{x}\sinh{y}

    ∵hxx+hyy=−cos⁡xsinh⁡y+cos⁡xsinh⁡y=0\because h_{xx}+h_{yy}=-\cos{x}\sinh{y}+\cos{x}\sinh{y}=0

    ∴h(x,y)=cos⁡xsinh⁡y\therefore h(x,y)=\cos{x}\sinh{y} is harmonic.