DerivatrveofcomplexfunctionExampleConsiderthefunctionf(z)=+2yif(z+△z) -f(2)Az(a+r)+2(y+△y)i-r-2yAr+Ayi1Ay=0,Ar→0Ar+2AyiAr+AyiAy→0.=02Sof(z)isnotdifferentiableinCQuQuQuOuBut u(c,y)=, (,y)=2y, their partial derivativesJr'oyor anddyexistandarecontinuous.Fromthis exampple,wehaveseenthata function whosefourpartialderivatives exist isnot sureanalyticFCV&IT13/45Fang Wang (Changsha Uni.of Sci &Tech)November5,2019
Derivative of complex functions Example Consider the function f(z) = x + 2yi f(z + ∆z) − f(z) ∆z = (x + ∆x) + 2(y + ∆y)i − x − 2yi ∆x + ∆yi = ∆x + 2∆yi ∆x + ∆yi → ( 1 ∆y = 0, ∆x → 0 2 ∆y → 0, ∆x = 0 . So f(z) is not differentiable in C. But u(x, y) = x, v(x, y) = 2y, their partial derivatives ∂u ∂x, ∂u ∂y , ∂v ∂x and ∂v ∂y exist and are continuous. From this exampple, we have seen that a function whose four partial derivatives exist is not sure analytic. Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 13 / 45
Letf(z)=u(α,y)+iv(r,y)inadomainD,if f(z)isdifferentiableatZo ED,thenf(z) -f(zo)f(z0)= lim :2→20z-20Let takethe special casethat z=r+iyo-Thenf(z)-f(zo)(r,yo)+i(r,yo)-u(ro,yo)+iu(ao,yo)Z-2OT-TOu(r,yo) -u(ro,yo) (r,yo) -v(ro,y0)T-TOT-rIn this case, as → 20. we have 2 → 20 and f() -f(20) ,2→f'(20)-20Thus both real and imaginary parts of the right sidemust convergeto alimit.From the definition of the partial derivatives this limit isauau+ir(r0.90)0r(r0.90)FCV&ITNovember.5,201914/45haUni.ofSci&Tech)angWan1C
Derivative of complex functions Let f(z) = u(x, y) + iv(x, y) in a domain D, if f(z) is differentiable at z0 ∈ D, then f 0 (z0) = limz→z0 f(z) − f(z0) z − z0 Let take the special case that z = x + iy0. Then f(z) − f(z0) z − z0 = u(x, y0) + iv(x, y0) − u(x0, y0) + iv(x0, y0) x − x0 = u(x, y0) − u(x0, y0) x − x0 + i v(x, y0) − v(x0, y0) x − x0 In this case, as x → x0, we have z → z0 and f(z) − f(z0) z − z0 → f 0 (z0). Thus both real and imaginary parts of the right side must converge to a limit. From the definition of the partial derivatives this limit is ∂u ∂x (x0,y0) + i ∂v ∂x (x0,y0) Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 14 / 45
Derivative of conlexfunctionsNext let z= Co + iy,then we similarly havef(z)-f(zo)u(ro,y)+iv(ro,y)-u(ro,yo)+iv(ro,yo)i(y-yo)z-20u(ro,y) -u(ro, yo) v(ro,y) -v(ro, yo)i(y-yo)y-yoAs y→ yo, weget z→ zo, andaduau10u+f'(zo) =dy(ro.yo)y l(ro,90)y l(ro.yo)dyI(ro.yo)Thus,since f'(co)exists and has the same value regardless of howzapproachingzo,wegetu.ovauOuf'(zo)+OrdraydyFCV&ITNovember 5, 201915/45aUni..of Sci&Tech)angWa
Derivative of complex functions Next let z = x0 + iy, then we similarly have f(z) − f(z0) z − z0 = u(x0, y) + iv(x0, y) − u(x0, y0) + iv(x0, y0) i(y − y0) = u(x0, y) − u(x0, y0) i(y − y0) + v(x0, y) − v(x0, y0) y − y0 As y → y0, we get z → z0, and f 0 (z0) = 1 i ∂u ∂y (x0,y0) + ∂v ∂y (x0,y0) = ∂v ∂y (x0,y0) − i ∂u ∂y (x0,y0) Thus, since f 0 (x0) exists and has the same value regardless of how z approaching z0, we get f 0 (z0) = ∂u ∂x + i ∂v ∂x = ∂v ∂y − i ∂u ∂y Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 15 / 45
DerivatrvXfunetinnOu.OuOudydrardyBy comparing real and imaginary parts of these equations, we deriveauuarOyduaudyOrwhich are called Cauchy-Riemann equations(C-R equations or C-Rcondition)FCV&ITaUni.ofSci&Tech)November.5,201916/45
Derivative of complex functions ∂u ∂x + i ∂v ∂x = ∂v ∂y − i ∂u ∂y By comparing real and imaginary parts of these equations, we derive ∂u ∂x = ∂v ∂y ∂u ∂y = − ∂v ∂x which are called Cauchy-Riemann equations (C-R equations or C-R condition). Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 16 / 45
DerivatrveofconplexfunctionsTakeAz=z-zo=A+iyDu=u(r,y)-u(ro,yo)Au = u(r, y) -v(ro, yo)f(z) -f(zo)2 - f(z0) = e(z) = pi(z) + ip2(△z)Z-20ByAu+iuf(z) -f(zo)u(r, y) -u(ro, yo) +i(v(r,y) -v(ro, yo))Ar+iAy(r-ro)+ i(y-yo)z-20we haveAu+iu=f(zo)(A+iy)+(pi+ip2)(Ar+iy)au.o(Ar+iy)+(pi+ip2)(Ar+iy)oroFCV&ITaUni.of Sci &Tech)November.5,201917/45
Derivative of complex functions Take ∆z = z − z0 = ∆x + i∆y ∆u = u(x, y) − u(x0, y0) ∆v = v(x, y) − v(x0, y0) f(z) − f(z0) z − z0 − f 0 (z0) = ε(∆z) = ρ1(∆z) + iρ2(∆z) By f(z) − f(z0) z − z0 = u(x, y) − u(x0, y0) + i(v(x, y) − v(x0, y0)) (x − x0) + i(y − y0) = ∆u + i∆v ∆x + i∆y we have ∆u + i∆v = f 0 (z0)(∆x + i∆y) + (ρ1 + iρ2)(∆x + i∆y) = ∂u ∂x + i ∂v ∂x (∆x + i∆y) + (ρ1 + iρ2)(∆x + i∆y) Fang Wang (Changsha Uni. of Sci & Tech) FCV & IT November 5, 2019 17 / 45