Complex Analysis- Lecture 16 - Analytic Function

EX:- Find f^ (Z) for all following:-
f(Z)=3Z^2-2
f^ (Z)=6Z
f(Z)=iZ^2-(1-i)Z
f^ (Z)=2iZ-(1-i)
f(Z)=(Z+1)/(Z-1)

f^ (Z)=((Z-1)(1)-(Z+1(1))/(Z-1)^2 =(Z-1-Z-1)/(Z-1)^2 =(-2)/(Z-1)^2

2.7 Analytic Function:-
Def:- We say that a function ?=f(Z) is analytic function at Z_0 if ?=f(Z) is differentiable function at Z_0 and all points in neighborhood of Z_0.
Def:- We said to be the function ?=f(Z) is entire function if it is analytic function in all points Zplane.
Def:- Let f(Z) is analytic function in some points of neighborhood of Z_0 except Z_0 , then Z_0 is called singular point.
Polynomial is entire function.
EX 1:- f(Z)=|Z|^2 is differentiable function at Z=0. H.W
EX 2:- Prove that ?=f(Z)=|Z|^2 is differentiable function at Z=0 but It isn’t analytic function at Z=0 .
Sol:-
??/?Z=(|Z+?Z|^2-|Z|^2)/?Z=((Z+?Z)(Z ?+?Z ? )-ZZ ?)/?Z=(ZZ ?+Z?Z ?+?ZZ ?+ZZ ?)/?Z
= Z (?Z ?)/?Z+Z ?+?Z ? *
If Z=0 , then ??/?Z=?Z ? , we get lim?(?Z?0)????/?Z?=lim?(?Z?0) ?Z
=lim??(?x?0@?y?0) (?x-?y)=lim?(?y?0)??[(0)-?y]=lim?(?x?0)??(-?y)=-(0)? ?
=0=f^ (0)
That is the derivative of function ?=f(Z) at Z=0 is exist and equal 0.
We will prove that the derivative of this function isn’t exist at neighborhood of? Z?_0.
If Z?0 and ?Z=?x then ?Z ?=?x by put it in * , we get
d?/dZ=Z?(?Z ?)/?Z+Z ?+?Z ?=Z??x/?x+Z ?+?x=Z+Z ?+?x
Thus lim?(?Z?0)??d?/dZ=? lim?(?x?0) (Z+Z ?+?x)=Z+Z ?+(0)
And if ?Z=i?y , then ?Z ?=i?y by put it in * ,we get
d?/dZ=Z?(?Z ?)/?Z+Z ?+?Z ?=Z?(-i?y)/i?y+Z ?+(i?y)=-Z+Z ?-i?y
Thus lim?(?Z?0)??d?/dZ=? lim?(?y?0)??[-Z+Z ?-i?y]=-Z+Z ?-i(0)=-Z+? Z ?
SinceZ?0, then the limit of derivative of function ?=f(Z) isn t unique, we get the function ?=f(Z)=|Z|^2 isn t differentiable function at neighborhood of Z?0 , so that f(Z)=|Z|^2 isn t analytic function at Z?0.
Theorem: - If the function ?=f(Z)=u(x,y)+iv(x,y) is analytic function in region D, then u and v are satisfied Cauchy-Riemann equations ?u/?x=?v/?y and ?u/?y=-?v/?x . If partial derivative in C-R equations (?u/?x,?v/?y,?u/?y,?v/?x) are continuous function in region D ,then C-R equations are sufficient to make the function ?=f(Z) is analytic function in region D .
EX 1:- Prove that f(Z)=Z^2+5iZ+3-i is entire ?
SOL:-
f(Z)=(x+iy)^2+5i(x+iy)+3-i
=(x^2+y^2 )+2xyi+5ix+5i^2 y+3-i
=(x^2-y^2-5y+3)+i(2xy+5x-1)
u=x^2-y^2-5y+3 & v=2xy+5x-1
u_x=2x ,u_y=-2y-5 ,v_x=2y+5 ,v_y=2x
Thus u_x=2x=v_y
? u?_y=-2y-5=-(2y+5)=-v_x
The C-R equations are satisfied, we see the partial derivative of u,v are continuous functions. So that f(Z) is an entire function by theorem C-R.
EX(2):- The function f(Z)=Z-iZ ? isn t analytic function , prove that?
SOL:-
f(Z)=Z-iZ ?=(x+iy)-i(x-iy)
=x+iy-ix+i^2 y=x+iy-ix-y=(x-y)+i(y-x)
u=x-y & v=y-x
u_x=1 ,? u?_y=-1 ,? v?_x=-1 ,? v?_y=1
By C-R equations, we get
u_x=? v?_y?u_x=1=? v?_y
? u?_y=-v_(x )?u_y=-1 but-v_(x )=-(-1)=1
Thus u_y?-v_(x )
The C-R equations aren’t satisfied, so that f(Z) isn t analytic function.
EX(3):- Prove that the function ?=f(Z)=|Z|^2=x^2+y^2 is satisfied
C-R equations at Z=0 but it isn’t analytic function at Z=0.
SOL:-since u=x^2+y^2 & v=0
u_x=2x , u_y=2y ,v_(x )=0 ,? v?_y=0
Put Z=0?x=0 ,y=0 then
u_x=z (0)=0=? v?_y
u_y=z(0)=0=-v_(x )
The function is satisfied C-R equation at Z=0 but we known from prior example that f(Z)=|Z|^2 isn t analytic function.