chain rule of the derivative?-

chain rule of the derivative?-
Statement ?-
suppose f is differentiable at z_0 and g is differentiable at f(z_0 ) then gof
is differentiable at z_0 .and (gof) ?(z_0 )=g ?(f(z_0 )).f ?(z_0)
example:-
find out the derivative of (z^3-2z^2+iz)^7
solution:
taking the help of chain rule me got
7.(z^3-2z^2+iz)^6.(3z^2-4z+i)
Cauchy-Riemann equation?-
statement:-
suppose f(z)=u+iv is differentiable at z=(x,y),then at (x,y)
the first order partial derivative of u and v must
exist ,u_x=v_y and u_y=?-v?_x must be true at (x,y),further.
f ?(z)=u_x+iv_x ,where the partial derivative are evaluated at (x,y).
these partial differentiable equation
i.e u_x=v_y and u_y=-v_x are called cauchy-Riemann equation
note?-
this theorem tells that for the differentiablilty of f at
z=(x,y) ,?u/?x-?v/?y=0 and ?u/?y+?v/?x=0
must be satisfied.
example:-
1.f(z)=x^2-y^2+i2xy
solution:-
in this case
u(x,y)=x^2-y^2
v(x,y)=2xy
then u_x=2x ,v_x=2y
u_y=-2y ,v_y=2x
so u_x=v_y and u_y=-v_x at each (x,y)? ? ,since by the previouse result
f (z)=u_x+?iv?_x=v_y-iu_y
=2x+i2y
=2(x+iy)=2z
equavelently ,f(z)=z^2 means f ?(z)=2z
2.suppose f(z)=|z|^2=x^2+y^2+i.0
solution:
u(x,y)=x^2+y^2
v(x,y)=0
while
u_x=2x ,u_y=2y
v_x=0 ,v_y=0
note that C.R.equation
i.e. u_x=v_y ,u_y=-v_x are true only at (0,0) further ,f ?(z) at (0,0) is
f ?(z)=u_x+iv_x=0 at (0,0)
3-f(z)=-sin??x cos??hy-i cos??x sin??hy ? ? ? ?
so for continuity & differentiability
here u(x,y)=-sinx coshy
v(x,y)=-cosx sinhy
u_x=-cosx coshy ,v_x=sinx sin?hy
u_y=-sinx sinhy ,v_y=-cosx coshy
u_x=v_y
u_y=-v_x
f ?(z)=u_x+iv_x=-cosx coshy+isinx sin?hy