Complex Analysis- Lecture 15 - Differentiation

2.5 Differentiation:-
Def:- Let a complex function f(Z) is defined in region R of the Z plan the derivative of function f(Z) at Z=Z_0 is defined as follows:
? f?^( ) (Z)=lim?(Z?Z_0 )??(f(Z)-f(Z_0 ))/(Z-Z_0 )?
If ?Z=Z-Z_0, then we can write above equation as follows:-
f^( ) (Z)=lim?(?Z?0)??(f(Z_0+?Z)-f(Z_0 ))/?Z?
EX 1:- If f(Z)=Z^2, then prove that f^( ) (Z)=2Z
SOL:-
f^( ) (Z)=lim?(?Z?0) ((Z+?Z)^2-Z^2)/?Z=lim?(?Z?0) (Z^2+2Z?Z+(?Z)^2-Z^2)/?Z
=lim?(?Z?0) (2Z?Z+(?Z)^2)/?Z=lim?(?Z?0) (2Z+?Z)
= lim?(?Z?0) (2Z)+lim?(?Z?0) (?Z)=2Z+0=2Z
Ex 2:- Prove that doesn t derivate of f(Z)=Z ? where I(z)?0.
Sol:-
(f(Z+?Z)-f(Z))/?Z=((Z+?Z) ?-Z ?)/?Z=(Z ?+?Z ?-Z ?)/?Z=(?Z ?)/?Z=?(x-iy)/?(x+iy) =(?x-i?y)/(?x+i?y)
lim?(?Z?0) (f(Z+?Z)-f(Z))/?Z=lim??(?x?0@?y?0)??(?x-i?y)/(?x+i?y)?=(0-i?y)/(0+i?y)=-1
lim?(?Z?0) (f(Z+?Z)-f(Z))/?Z=lim??(?x?0@?y?0)??(?x-i?y)/(?x+i?y)?=(?x-i(0))/(?x-i(0))=?x/?x=1
We find two limits that is we can t find this limits
lim?(?Z?0) (f(Z+?Z)-f(Z))/?Z
Thus the derivative of f(Z) isn t exist.
Theorem: - If f(Z) is differentiable function then f(Z)is continuous function . The inverse of this theorem is false.
Def:-Let ?=f(Z), then f^ (Z)=lim?(?Z?0) ??/?Z where
??=f( Z_0+?Z)-f(Z_0 )
Z
?-?_0
If f(Z) , f^ (Z) is continuous in regionD. Then ??=f^ (Z)?Z+??Z where ?Z?0 as ??0. And If we write ?Z=dZ then f^ (Z)dZ is called differential of f(Z). We can write it as df=f^ (Z)dZ .
That is d?/dZ=f^ (Z).
NOTES:-
The symbol d/dZ is differential operator.
If ?=f(Z) then
The first derivative of ? is d/dZ (?)=d?/dZ=f^ (Z)
The second derivative of ? is d^2/(dZ^2 ) (?)=d/dZ (d?/dZ)= (d^2 ?)/(dZ^2 )=f^ (Z)
The nth derivative of ? is d^n/(dZ^n )(?)
2.6 Rules for Differentiation:-
Let f(Z),g(Z) are differentiable function at Z , then
d/dZ {f(Z)±g(Z)}=d/dZ f(Z)±d/dZ g(Z)=f (Z)±g (Z)
d/dZ {cf(Z)}=c d/dZ f(Z)=cf^ (Z) Where c is any constant.
d/dZ {f(Z)g(Z)}=f(Z) d/dZ g(Z)+g(Z)d/dZ f(Z)
= f(Z) g^ (Z)+g(Z)f (Z)
d/dZ {f(Z)/(g(Z))}=(g(Z) d/dZ f(Z)-f(Z) d/dZ g(Z))/{g(Z)}^2 =(g(Z) f^ (Z) -f(Z)g (Z))/{g(Z)}^2
If g(Z)?0

If ?=f(k) where k=g(Z),then
d?/dZ=d?/dk?dk/dz
and if ?=f(k),k=g(M),M=h(Z),then
d?/dZ=d?/dk?dk/dM?dM/dZ
If ?=f(Z),then Z=f^(-1) (?), we get
d?/dZ=1/(dZ/d?)?f^ (Z)=1/((f^(-1) )^ (Z) )
If Z=f(t) and ?=g(t) where t is a parameter, then
d?/dZ=(d?/dt)/( dZ/dt)=(g^ (t))/(f^ (t))
We have also
d{f(Z)±g(Z) }=df(Z)±dg(Z)=f^ (Z)dZ±g^ (Z)dZ
=(f^ (Z) )±g^ (Z)dZ
d{f(Z)g(Z) }=f(Z)dg(Z)+g(Z)df(Z)
=f(Z) g^ (Z)dZ+g(Z) f^ (Z)dZ
={f(Z) g^ (Z)+g(Z) f^ (Z) }dZ