Mathematical Methods for Physics and Engineering - Matematica.NET

24.8 COMPLEX INTEGRALSyRtyyC 1 C 2Rt s =1t =0x −RR x −RR xC 3biRC3a(a) (b) (c)Figure 24.9details.Different paths for an integral of f(z) =z −1 . See the text forThis very important result will be used many times later, and the following shouldbe carefully noted: (i) its value, (ii) that this value is independent of R.In the above example the contour was closed, and so it began and ended atthe same point in the Argand diagram. We can evaluate complex integrals alongopen paths in a similar way.◮Evaluate the complex integral of f(z) =z −1 along the following paths (see figure 24.9):(i) the contour C 2 consisting of the semicircle |z| = R in the half-plane y ≥ 0,(ii) the contour C 3 made up of the two straight lines C 3a and C 3b .(i) This is just as in the previous example, except that now 0 ≤ t ≤ π. With this change,we have from (24.35) or (24.36) that∫dz= πi. (24.37)C 2z(ii) The straight lines that make up the countour C 3 may be parameterised as follows:C 3a , z =(1− t)R + itR for 0 ≤ t ≤ 1;C 3b , z = −sR + i(1 − s)R for 0 ≤ s ≤ 1.With these parameterisations the required integrals may be written∫ ∫dz 1∫C 3z = −R + iR1R + t(−R + iR) dt + −R − iRds. (24.38)iR + s(−R − iR)0If we could take over from real-variable theory that, for real t, ∫ (a+bt) −1 dt = b −1 ln(a+bt)even if a and b are complex, then these integrals could be evaluated immediately. However,to do this would be presuming to some extent what we wish to show, and so the evaluation8470