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- .MCD 20000 0
- .CMD PLOTFORMAT logs=0,0 subdivs=1,1 size=5,15 type=l
- .CMD FORMAT rd=d ct=6 im=i et=3 zt=15 pr=3 mass length time charge
- .CMD SET ORIGIN 0
- .CMD SET TOL 0.001000
- .CMD MARGIN 0
- .CMD LINELENGTH 78
- .CMD SET PRNCOLWIDTH 8
- .CMD SET PRNPRECISION 4
- .TXT 1 0 1 28
- a1,27,25,26
- COMPLEX CONTOUR INTEGRALS
- .TXT 2 28 3 37
- a3,37,33,87
- This document shows the general
- technique for evaluating a complex
- contour integral.
- .TXT 3 -28 4 15
- a4,15,13,46
- Start with a
- complex
- function to
- evaluate:
- .EQN 2 28 3 11
- f(z):1/z
- .TXT 5 -28 4 17
- a4,16,15,47
- Choose a
- parameterized
- path in the z
- plane.
- .EQN 1 29 1 28
- z(t):cos(t)+1i*sin(t)
- .EQN 2 0 2 14
- t.start:0
- .EQN 0 21 2 14
- t.end:2*π
- .TXT 4 -50 2 35
- a2,34,32,60
- Then the integral of f over the
- path can be defined as ...
- .EQN 3 29 6 43
- (t.start&t.end`f(z(t))*(t"z(t))&t)=?
- .TXT 7 1 3 43
- a3,43,41,87
- This is 2πi, not zero, indicating a pole
- within the circular path described by
- z(t).
- .EQN 4 0 2 12
- TOL~10^-6
-