Mathematical Methods for Physics and Engineering - Matematica.NET

24.2 THE CAUCHY–RIEMANN RELATIONS◮Show that the function f(z) =1/(1 − z) is analytic everywhere except at z =1.Since f(z) is given explicitly as a function of z, evaluation of the limit (24.1) is somewhateasier. We find[ ]f(z +∆z) − f(z)f ′ (z) = lim∆z→0 ∆z[ ( 1 1= lim∆z→0 ∆z 1 − z − ∆z − 1 )]1 − z= lim∆z→0[1(1 − z − ∆z)(1 − z)]=1(1 − z) 2 ,independently of the way in which ∆z → 0, provided z ≠ 1. Hence f(z) is analyticeverywhere except at the singularity z =1.◭24.2 The Cauchy–Riemann relationsFrom examining the previous examples, it is apparent that for a function f(z)to be differentiable and hence analytic there must be some particular connectionbetween its real and imaginary parts u and v.By considering a general function we next establish what this connection mustbe. If the limit[ ]f(z +∆z) − f(z)L = lim(24.2)∆z→0 ∆zis to exist and be unique, in the way required for differentiability, then any twospecific ways of letting ∆z → 0 must produce the same limit. In particular, movingparallel to the real axis and moving parallel to the imaginary axis must do so.This is certainly a necessary condition, although it may not be sufficient.If we let f(z) =u(x, y)+iv(x, y) and∆z =∆x + i∆y then we havef(z +∆z) =u(x +∆x, y +∆y)+iv(x +∆x, y +∆y),and the limit (24.2) is given by[ u(x +∆x, y +∆y)+iv(x +∆x, y +∆y) − u(x, y) − iv(x, y)L = lim∆x, ∆y→0∆x + i∆yIf we first suppose that ∆z is purely real, so that ∆y = 0, we obtain[ u(x +∆x, y) − u(x, y)L = lim+ i∆x→0 ∆xv(x +∆x, y) − v(x, y)∆x].]= ∂u∂x + i ∂v∂x , (24.3)provided each limit exists at the point z. Similarly, if ∆z is taken as purelyimaginary, so that ∆x = 0, we find[ ]u(x, y +∆y) − u(x, y) v(x, y +∆y) − v(x, y)L = lim+ i = 1 ∂u∆y→0 i∆yi∆yi ∂y + ∂v∂y . (24.4)827