Lecture 17 Appendix B (analytic functions and contour integrals)

Transcription

1 Lecture 7 Appendix B (analytic functions and contour integrals) Ex 4.:8 We want to consider the analyticity properties (CR) of the square root function In[]:= f8@z_d := Sqrt@zD Now write this function in terms of its real and imaginary parts (carefully telling Mathematica that x,y are real) In[]:= U8@x_, y_d := Refine@Re@ComplexExpand@f8@x ä yddd, Element@x V8@x_, y_d := Refine@Im@ComplexExpand@f8@x ä yddd, Element@x In[3]:= U8@x, yd Out[3]= In[4]:= Out[4]= In[5]:= Out[5]= Ix y M 4 Arg@x ä ydf CosB FullSimplifyBIx y M Ix y M 4 y, RealsDD; y, RealsDD 4 CosB ArcTan@x, ydff CosB ArcTan@x, ydf V8@x, yd Ix y M 4 Arg@x ä ydf SinB Note that the arg function is the familiar arctan. Then we check the C R relations In[6]:= D@U8@x, yd, xd D@V8@x, yd, yd Out[6]= x CosA Arg@x ä yde Ix y M y SinA Arg@x ä yde Ix y M 3 4 ä Ix y M 4 Ix y M CosB SinB Arg@x ä ydf ä yd Arg@x ä ydf ä yd Mathematica doesn't like to take the derivative of the Arg function so try replacing it with the arctan In[7]:= In[8]:= U8@x_, y_d := Ix y M 4 CosB ArcTan@x, ydf V8@x_, y_d := Ix y M 4 SinB ArcTan@x, ydf

2 Lec7_8_9 App B.nb In[9]:= yd, xd yd, yd Out[9]= In[]:= yd, yd yd, xd Out[]= So the square root is analytic everywhere except at the branch point at the origin z =, where the above expressions diverge, although this feature is more easier to see in the analytic analysis. It is also unfortunate that Mathematica uses the arg function and then does not want to take derivatives. Ex 4.: 9 We want to consider the analyticity properties (CR) of the log function In[]:= f9@z_d := Log@zD Now write this function in terms of its real and imaginary parts (carefully telling Mathematica that x,y are real) In[]:= U9@x_, y_d := Refine@Re@ComplexExpand@f9@x ä yddd, Element@x V9@x_, y_d := Refine@Im@ComplexExpand@f9@x ä yddd, Element@x In[3]:= U9@x, yd Out[3]= In[4]:= ReALogAx y EE V9@x, yd Out[4]= y, RealsDD; y, RealsDD Arg@x ä yd ImALogAx y EE So Mathematica is not certain that the log is real try In[5]:= U9@x_, y_d := Refine@Re@ComplexExpand@f9@x ä yddd, Element@x y Log@x ^ y ^ D, RealsDD; V9@x_, y_d := Refine@Im@ComplexExpand@f9@x ä yddd, Element@x y Log@x ^ y ^ D, RealsDD In[6]:= U9@x, yd Out[6]= In[7]:= Out[7]= LogAx y E V9@x, yd Arg@x ä yd Note this last expression is the familiar arctan. Then we check the C R relations In[8]:= D@U9@x, yd, xd D@V9@x, yd, yd x Out[8]= x y ä ä yd Mathematica doesn't like to take the derivative of the Arg function so try

3 Lec7_8_9 App B.nb In[9]:= y_d := yd In[]:= yd, xd yd, yd Out[]= In[]:= Out[]= 3 D@U9@x, yd, yd D@V9@x, yd, xd So the log is analytic everywhere except at the branch point at the origin z =, where the above expressions diverge. Ex 4.:3 We want to consider the analyticity properties (CR) of the function In[]:= f3@z_d := z which we expect to be analytic except at the origin. Now write this function in terms of its real and imaginary parts (carefully telling Mathematica that x,y are real) In[3]:= U3@x_, y_d := Refine@Re@ComplexExpand@f3@x ä yddd, Element@x V3@x_, y_d := Refine@Im@ComplexExpand@f3@x ä yddd, Element@x In[4]:= U3@x, yd Out[4]= y ImB x y In[5]:= V3@x, yd Out[5]= x ImB x y F x ReB x y F y ReB x y y, RealsDD; y, RealsDD F F So Mathematica is not certain that /(x^y^) is real try In[6]:= U3@x_, y_d := RefineBRe@ComplexExpand@f3@x ä yddd, ElementBx y, RealsFF; x y V3@x_, y_d := RefineBIm@ComplexExpand@f3@x ä yddd, ElementBx y, RealsFF x y In[7]:= U3@x, yd x Out[7]= x In[8]:= y V3@x, yd y Out[8]= x y Then we check the C R relations

4 4 Lec7_8_9 App B.nb In[9]:= yd, xd yd, yd x Out[9]= y Ix y M In[3]:= Out[3]= In[3]:= Out[3]= Ix y M Simplify@%D x y D@U3@x, yd, yd D@V3@x, yd, xd So /z is analytic everywhere except at the origin where the above expressions diverge. Ex 4.:34 We want to consider the analyticity properties (CR) of the log function In[3]:= f34@z_d := Log@ zd whose series expansion is In[33]:= Series@f34@zD, 8z,, 4<D z Out[33]= z z3 z4 3 O@zD5 4 Now write this function in terms of its real and imaginary parts (carefully telling Mathematica that x,y are real) In[34]:= U34@x_, y_d := Refine@Re@ComplexExpand@f34@x ä yddd, Element@x V34@x_, y_d := Refine@Im@ComplexExpand@f34@x ä yddd, Element@x In[35]:= U34@x, yd Out[35]= In[36]:= ReALogAH xl y EE V34@x, yd Out[36]= y, RealsDD; y, RealsDD Arg@ x ä yd ImALogAH xl y EE So Mathematica is not certain that the log is real try In[37]:= U34@x_, y_d := Refine@Re@ComplexExpand@f34@x ä yddd, Element@x y Log@H xl ^ y ^ D, RealsDD; V34@x_, y_d := Refine@Im@ComplexExpand@f34@x ä yddd, Element@x y Log@H xl ^ y ^ D, RealsDD In[38]:= U34@x, yd Out[38]= In[39]:= Out[39]= LogAH xl y E V34@x, yd Arg@ x ä yd

5 Lec7_8_9 App B.nb 5 Note this last expression is the familiar arctan. Then we check the C R relations In[4]:= D@U34@x, yd, xd D@V34@x, yd, yd x Out[4]= H xl y So try ä x ä yd In[4]:= V34@x_, y_d := ArcTan@ x, yd In[4]:= D@U34@x, yd, xd D@V34@x, yd, yd Out[4]= In[43]:= Out[43]= D@U34@x, yd, yd D@V34@x, yd, xd So the log is analytic everywhere except at the branch point at the point z =, where the above expressions diverge. Ex 4.:48 This exercise is really best suited for an analytic analysis. Ex 4.3:7 Here we try to use Mathematica to directly evaluate (closed path or contour) integrals of the same integrand around circles with different radii. We have Sin@zD In[44]:= F7@z_D := zπ The contours and the pole at z = Π/ look like

6 6 Lec7_8_9 App B.nb In[45]:= 8Thick, <, D, Thick, Dashed, <, D, <,.5D, 8,.<<D, 8,.<<D<, Axes TrueD Out[45]= Now change to polar representation, z = RãäΘ, dz = ärãäθ dθ. So we can try to perform the integrals directly For R = In[46]:= Out[46]= I7A = IntegrateAF7Aãä Θ E Iä ãä Θ M, 8Θ,, Π<E As expected (from Cauchy) it vanishes as there is not pole inside the contour. For R = In[47]:= Out[47]= I7B = IntegrateAF7A ãä Θ E I ä ãä Θ M, 8Θ,, Π<E äπ This agrees with our analytic results. For fun we can also try contour integrals with NIntegrate using square contours (so easier to specify) that have the same poles enclosed.

7 Lec7_8_9 App B.nb In[48]:= <,.5D, 8,.<<D, 8,.<<D, <, 8, <D, DashedDD, <, 8, <D<, Axes TrueD Out[48]= In[49]:= 8z, ä, ä, ä, ä, ä<d NIntegrate::slwcon : Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is, highly oscillatory integrand, or WorkingPrecision too small. NIntegrate::ncvb : NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in z near 8z< = ä<. NIntegrate obtained ä and `*^4 for the integral and error estimates. Out[49]= In[5]:= Out[5]= In[5]:= Out[5]= In[5]:= Out[5]= ä Chop@%D NIntegrate@F7@zD, 8z, ä, ä, ä, ä, ä<d ä N@ΠD

8 8 Lec7_8_9 App B.nb Ex 4.3:3 We want to consider the contour integral with integrand zd In[53]:= := Hz ^ 4 Since the contour is a square, probably the simplest approach in Mathematica is to use In[54]:= Out[54]= In[55]:= Out[55]= NIntegrate@f3@zD, 8z, ä, ä, ä, ä, ä<d ä Chop@%D ä where we notice that analytic answer is the same. In[56]:= N@7 ΠD Out[56]= 6.95 We can also use Mathematica to look for the residue z = Log[] In[57]:= Out[57]= Residue@f3@zD, 8z, Log@D<D 36 Or Π ä times 36 gives 7 Π ä. Ex 4.6: 3 Consider the function Sin@zD In[58]:= f3@z_d := z^4 To find the (Laurent) series we can just use Series, which includes the singular terms In[59]:= Series@f3@zD, 8z,, 3<D Out[59]= z3 6z z3 z O@zD4 54 which agrees with our analytic result and where the first term gives the residue. Note also that we can find the residue from In[6]:= Residue@f3@zD, 8z, <D Out[6]= 6 Ex 4.6: 5 Consider the function

9 Lec7_8_9 App B.nb In[6]:= := z^ To find the (Laurent) series we can just use Series, which includes the singular terms In[6]:= 8z,, 3<D ã Out[6]= Hz L ã 4 8 ã Hz L 48 ã Hz L 96 ã Hz L3 O@z D4 which agrees with our analytic result and where the first term gives the residue. Note also that we can find the residue from In[63]:= Residue@f5@zD, 8z, <D ã Out[63]= Ex 4.7: We want to consider an angular integral with the contour equal a circle. Proceeding directly we have In[64]:= IntegrateB 3 5 Sin@ΘD Π Out[64]=, 8Θ,, Π<F 6 This is a pretty easy check of the analytic result. Ex 4.7:3 We want to consider an integrand with many poles. First we note that Mathematica can simply do the integral. x^ In[65]:= I3 = IntegrateB Π Out[65]= H x ^ 4L Hx ^ 9L, 8x,, <F Alternatively if we wanted to use Cauchy we could use Mathematica to find the residues. In[66]:= f3@z_d := z^ H z ^ 4L Hz ^ 9L Closing above In[67]:= Residue@f3@zD, 8z, ä<d ä Out[67]= In[68]:= Residue@f3@zD, 8z, 3 ä<d 3ä Out[68]= 9

10 Lec7_8_9 App B.nb Or closing below in CW direction so overall minus sign In[69]:= 8z, ä<d ä Out[69]= In[7]:= 8z, 3 ä<d 3ä Out[7]= Ex 4.7:9 The first Mathematica check is just to do the integral. Cos@ xd In[7]:= I9 = IntegrateB Π Out[7]= 54 ã3 H4 x ^ 9L ^, 8x,, <F But we can also use it to find the residues In[7]:= In[73]:= f9@z_d := Exp@ z äd H4 z ^ 9L ^ ResidueBf9@zD, :z, 3 ä>f ä Out[73]= 8 ã3 This agrees with what we found analytically after some arithmetic. Ex 4.7:33 Now consider an integrand with a square root. Again Mathematica will proceed directly. x In[74]:= I33 = IntegrateB x Π Out[74]=, 8x,, <F In this case, using Mathematica is considerably easier than proceeding analytically.