Presentation is loading. Please wait.

Presentation is loading. Please wait.

講者: 許永昌 老師 1. Contents Conformal Mapping Mappings Translation Rotation Inversion Branch Points and Multivalent Functions 2.

Similar presentations


Presentation on theme: "講者: 許永昌 老師 1. Contents Conformal Mapping Mappings Translation Rotation Inversion Branch Points and Multivalent Functions 2."— Presentation transcript:

1 講者: 許永昌 老師 1

2 Contents Conformal Mapping Mappings Translation Rotation Inversion Branch Points and Multivalent Functions 2

3 Conformal mapping ( 請預讀 P368~P370) 3

4 Conformal mapping (continue) Based on Cauchy-Riemann conditions, we get  2 u=0=  2 v,  u   v=0. They are orthogonal to each other  The curves u=constant and v=constant are orthogonal to each other. Example: w=z 2 =(x 2 -y 2 )+2ixy Code: z2_uv.mz2_uv.m y=sqrt(x 2 -u), y=v/(2x) Contour u= x 2 -y 2, v=2xy. 4 Proper rotation

5 Conformal Mapping (final) The mapping of w=z 2. From these figures, you will find that the contour lines of y=C and y=-C are the same in w-plane. Reason: z=re i  and z’=re i  i . z 2 =r 2 e i2  =z’ 2. Therefore, it has a two-to-one correspondence. 5

6 Mappings ( 請預讀 P360~P363) Linear Transformation: Translation: w=z+z 0. Rotation: w=cz=(r  r c )e i(  c ). Nonlinear Transformation: Inversion: w=1/z=1/r  e  i . … Code: mappings.mmappings.m 想像 w=z‘ 與 z 畫在同一個座標系 6

7 Exercise Prove that w=1/z will map a straight line in z-plane into a circle cross w=0. Try to add “z=z*(1+1i);x=real(z);y=imag(z);” into the code mappings.m to see the result. 7

8 Multivalent functions and Branch Points ( 請預讀 P363~P367) 8 Hint: Restrict the allowed range of  ’.

9 Multivalent functions and Branch Points (continue) 9 * 莊 ( 土斤 ) 泰,張南岳,復變函數

10 Multivalent functions and Branch Points (final) 10

11 Homework

12 Nouns 12


Download ppt "講者: 許永昌 老師 1. Contents Conformal Mapping Mappings Translation Rotation Inversion Branch Points and Multivalent Functions 2."

Similar presentations


Ads by Google