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CHAPTER 3 ANALYTIC FUNCTION
Defini ti on f un c d efine d in a ne i g hbor hoo d of z is sai d to be d iff at z iff lim
f ( z h ) f ( z )
h
exists uniq uely ,
w hen h thr ou g h co m p lex v alues. Th is li m is called d eriv at ive of f at z i s d en ote d b y f ( z )
? f ( z ) lim
f ( z h ) f ( z )
h
Thm 3.1 Let f = u+iv be diff at z. Then f satisfy C-R eq
n
f y =i f x & converse is not true.
Proof: Let the fun
c
f is diff at z.
lim
f ( z h ) f ( z )
h
exist uniq uely as h along any d ir
Let z = x + i y & h=h 1+i h 2
ake h along r ea l axis, then h 2= 0
? f ( z ) lim
f ( x h
h , y ) f ( x , y )
h h
= f x … … … … … … … … … … … (i)
ake h along imag in ary axis then h 1
? f
( z ) lim
f ( x , y h ) f ( x , y )
h
lim
f ( x , y h ) f ( x , y )
i h
i
lim
f ( x , y h ) f ( x , y )
h
i
f
… … … ( ii )
? f r om ( i) ( ii )
i f x = f y
but converse of the above thm is not true.
i.e. f x & f y may exists in a nbhd of z & may satisfy C-R eq
n
f y=i f x but the fun
c
f is not diff at z.
E xample f ( x ) {
xy ( x iy )
x
y
, if z , if z
Now C-R eq
n
at z=0
f
lim
f ( h , ) f ( , )
h
lim
h
lim
h
(
* f or m
lim
by L ’Hop it a l r ule
f
lim
f ( , h ) f ( , )
h
lim
f ( , h ) f ( , )
h
lim
h
lim
h
(
* in d et erm in at e f or m
lim
H en ce C R eq n sat isfie d b ut not d iff at z
Consider,
lim
f ( z ) f ( )
z lim
xy ( x iy )
x
y
x i y
lim
xy
x
y
lim
( , ) ( , )
xy
x
y
Taking along the path y=mx as x , y
lim
( , ) ( , )
x mx
x
( mx )
lim
( , ) ( , )
x mx
x
( m
)
h =h 1+0 i
h real ax i s t h en h 1
i f h al ong im ag i n ary a x i s t he n h 2
also h= 0+ih 2
Page 2
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CHAPTER 3 ANALYTIC FUNCTION
Defini ti on f un c d efine d in a ne i g hbor hoo d of z is sai d to be d iff at z iff lim
f ( z h ) f ( z )
h
exists uniq uely ,
w hen h thr ou g h co m p lex v alues. Th is li m is called d eriv at ive of f at z i s d en ote d b y f ( z )
? f ( z ) lim
f ( z h ) f ( z )
h
Thm 3.1 Let f = u+iv be diff at z. Then f satisfy C-R eq
n
f y =i f x & converse is not true.
Proof: Let the fun
c
f is diff at z.
lim
f ( z h ) f ( z )
h
exist uniq uely as h along any d ir
Let z = x + i y & h=h 1+i h 2
ake h along r ea l axis, then h 2= 0
? f ( z ) lim
f ( x h
h , y ) f ( x , y )
h h
= f x … … … … … … … … … … … (i)
ake h along imag in ary axis then h 1
? f
( z ) lim
f ( x , y h ) f ( x , y )
h
lim
f ( x , y h ) f ( x , y )
i h
i
lim
f ( x , y h ) f ( x , y )
h
i
f
… … … ( ii )
? f r om ( i) ( ii )
i f x = f y
but converse of the above thm is not true.
i.e. f x & f y may exists in a nbhd of z & may satisfy C-R eq
n
f y=i f x but the fun
c
f is not diff at z.
E xample f ( x ) {
xy ( x iy )
x
y
, if z , if z
Now C-R eq
n
at z=0
f
lim
f ( h , ) f ( , )
h
lim
h
lim
h
(
* f or m
lim
by L ’Hop it a l r ule
f
lim
f ( , h ) f ( , )
h
lim
f ( , h ) f ( , )
h
lim
h
lim
h
(
* in d et erm in at e f or m
lim
H en ce C R eq n sat isfie d b ut not d iff at z
Consider,
lim
f ( z ) f ( )
z lim
xy ( x iy )
x
y
x i y
lim
xy
x
y
lim
( , ) ( , )
xy
x
y
Taking along the path y=mx as x , y
lim
( , ) ( , )
x mx
x
( mx )
lim
( , ) ( , )
x mx
x
( m
)
h =h 1+0 i
h real ax i s t h en h 1
i f h al ong im ag i n ary a x i s t he n h 2
also h= 0+ih 2
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lim
m
m
m
m
f or d iff v alue of m we f in d d iff erent lim
lim is not uniq ue .
? f is n ot d iff at z .
Note (i) f is said to be analytic at z iff f is diff in the nbhd of z.
(ii) f is said to be analytic in a set S if f is diff in an open set containing S.
3.6 Proposition
If f = u+iv be analytic in a region D and u is constant, and then f is constant.
Proof let f be analytic in a region D.
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+iv x)
v y = u x
u y v x …… ( i)
given: u consta nt
u x = 0, u y =0
v y = 0, v x =0 by (i)
v consta nt
f is const an t.
Result let f = u+i v be analytic in a region D. Let v is constant, and then f is constant.
Proof let f be analytic in a region D
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+i v x)
u y v x
v y = u x
now, It is given that v is constant
p artia l d eriv at ive of v mu st b e zer o
v x=v y =0
u x = u y =0 b y C R eq
n
i.e. partial derivative of u(x, y) are zero
u is consta nt hen c e f is const an t
3.7 Proposition
If f =u+i v be analytic in a region D and if |f| be constant there, then f is constant.
Sol: let f = u+iv
? |f| = v u
v
= constant
u
2
+v
2
c ( say) …… …… . . ( i)
Case 1 if c = 0
? u
2
+v
2
= 0
u
2
= 0 & v
2
= 0
u = 0 & v = 0
u i v = 0
f = 0 f is const an t
Case 2 if c
Diff partially of eq
n
(i) w.r . t. ‘x ’ ‘y ’
2 u u x + 2 v v x = 0
& 2 u u y + 2 v v y = 0
u u x + v v x = 0
& u u y + v v y = 0
u v y + v v x …… . . ( ii )
u v x + v v y …… …. . ( i ii )
Multiply eq
n
(ii) by v, eq
n
(iii) by u and add
u v y +v v x = 0
u ( v x) + v v y =0
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CHAPTER 3 ANALYTIC FUNCTION
Defini ti on f un c d efine d in a ne i g hbor hoo d of z is sai d to be d iff at z iff lim
f ( z h ) f ( z )
h
exists uniq uely ,
w hen h thr ou g h co m p lex v alues. Th is li m is called d eriv at ive of f at z i s d en ote d b y f ( z )
? f ( z ) lim
f ( z h ) f ( z )
h
Thm 3.1 Let f = u+iv be diff at z. Then f satisfy C-R eq
n
f y =i f x & converse is not true.
Proof: Let the fun
c
f is diff at z.
lim
f ( z h ) f ( z )
h
exist uniq uely as h along any d ir
Let z = x + i y & h=h 1+i h 2
ake h along r ea l axis, then h 2= 0
? f ( z ) lim
f ( x h
h , y ) f ( x , y )
h h
= f x … … … … … … … … … … … (i)
ake h along imag in ary axis then h 1
? f
( z ) lim
f ( x , y h ) f ( x , y )
h
lim
f ( x , y h ) f ( x , y )
i h
i
lim
f ( x , y h ) f ( x , y )
h
i
f
… … … ( ii )
? f r om ( i) ( ii )
i f x = f y
but converse of the above thm is not true.
i.e. f x & f y may exists in a nbhd of z & may satisfy C-R eq
n
f y=i f x but the fun
c
f is not diff at z.
E xample f ( x ) {
xy ( x iy )
x
y
, if z , if z
Now C-R eq
n
at z=0
f
lim
f ( h , ) f ( , )
h
lim
h
lim
h
(
* f or m
lim
by L ’Hop it a l r ule
f
lim
f ( , h ) f ( , )
h
lim
f ( , h ) f ( , )
h
lim
h
lim
h
(
* in d et erm in at e f or m
lim
H en ce C R eq n sat isfie d b ut not d iff at z
Consider,
lim
f ( z ) f ( )
z lim
xy ( x iy )
x
y
x i y
lim
xy
x
y
lim
( , ) ( , )
xy
x
y
Taking along the path y=mx as x , y
lim
( , ) ( , )
x mx
x
( mx )
lim
( , ) ( , )
x mx
x
( m
)
h =h 1+0 i
h real ax i s t h en h 1
i f h al ong im ag i n ary a x i s t he n h 2
also h= 0+ih 2
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lim
m
m
m
m
f or d iff v alue of m we f in d d iff erent lim
lim is not uniq ue .
? f is n ot d iff at z .
Note (i) f is said to be analytic at z iff f is diff in the nbhd of z.
(ii) f is said to be analytic in a set S if f is diff in an open set containing S.
3.6 Proposition
If f = u+iv be analytic in a region D and u is constant, and then f is constant.
Proof let f be analytic in a region D.
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+iv x)
v y = u x
u y v x …… ( i)
given: u consta nt
u x = 0, u y =0
v y = 0, v x =0 by (i)
v consta nt
f is const an t.
Result let f = u+i v be analytic in a region D. Let v is constant, and then f is constant.
Proof let f be analytic in a region D
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+i v x)
u y v x
v y = u x
now, It is given that v is constant
p artia l d eriv at ive of v mu st b e zer o
v x=v y =0
u x = u y =0 b y C R eq
n
i.e. partial derivative of u(x, y) are zero
u is consta nt hen c e f is const an t
3.7 Proposition
If f =u+i v be analytic in a region D and if |f| be constant there, then f is constant.
Sol: let f = u+iv
? |f| = v u
v
= constant
u
2
+v
2
c ( say) …… …… . . ( i)
Case 1 if c = 0
? u
2
+v
2
= 0
u
2
= 0 & v
2
= 0
u = 0 & v = 0
u i v = 0
f = 0 f is const an t
Case 2 if c
Diff partially of eq
n
(i) w.r . t. ‘x ’ ‘y ’
2 u u x + 2 v v x = 0
& 2 u u y + 2 v v y = 0
u u x + v v x = 0
& u u y + v v y = 0
u v y + v v x …… . . ( ii )
u v x + v v y …… …. . ( i ii )
Multiply eq
n
(ii) by v, eq
n
(iii) by u and add
u v y +v v x = 0
u ( v x) + v v y =0
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(u
2
+v
2
)v y = 0
v y ( u
2
+v
2
)
Similarly, u y = 0
? b y C R eq
n
u x v x = 0
? u x = u y = v x = v y = 0
u consta nt v co nst an t
f is const an t
LMV If a fun
c
f is defined on a [a, a+h] s.t.
(i) f continuous on [a, a+h]
(ii) f d iff ( a, a h) the n ? ( , ) s. t.
f(a+h) – f(a) = h f
( a h)
for 2 variable
f(a+h, b) – f(a, b) = h f x ( a h , b )
f(a, b+h) – f(a, b) = h f y ( a, b h)
Thm 3.2 Let f x & f y exists in a nbhd of z, Let f x & f y b e ct s at z , le t f sat isfy C R eq
n
f y = i f x
i.e. u x = v y
u y v x
proof let f = u+i v , z = x +i y
we have to prove
lim
f ( z h ) f ( z )
h
f
( z )
L et h i z = x+i y
z h x i ( y )
consider,
f ( z h ) f ( z )
h
u ( x , y ) i v ( x , y ) - ,u ( x , y ) i v ( x , y ) -
i
u ( x , y ) u ( x , y ) - i , v ( x , y ) v ( x , y ) -
i … . ( i )
Now,
u( x , y ) u( x, y) u( x , y ) u ( x , y ) u ( x , y )
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
–u(x, y)
,u( x , y ) u( x, y )- ,u( x, y ) u ( x, y ) -
u x ( x 1 , y ) u y(x, y 2 )
w her e 1 , 2 by L V …… ( ii)
u x(z 1 ) u y(z 2)
Where z 1 ( x 1 ) i (y )
z 2 x i ( y 2 )
&|z 1 z | | z 2 z | as h or ,
Similarly,
v ( x , y ) v ( x, y) v x(z 3 ) v y(z 4) …… …. . ( iii)
where |z 3 z | | z 4 z | as h
from (i) , (ii) & (iii)
f ( z h ) f ( z )
h
u
( z
) u
( z
) - i , v
( z
) v
( z )
i
, u
( z
) i v
( z
) - ,u
( z
) i v
( z
)
i … ( iv )
Now, f x(z) = u x(z)+i v x(z)
,u x ( z ) i v x ( z ) - i i
,u
( z ) i v
( z ) - , i u
( z ) v
( z ) -
i
Then f is diff at z & f
(z) = f x(z)
suppose f is analytic in a region R, if one of
the foll
n
hold
(1) Im f is constant on
(2) Re f is constant on
(3) |f| is constant on
Then f is constant on
Page 4
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CHAPTER 3 ANALYTIC FUNCTION
Defini ti on f un c d efine d in a ne i g hbor hoo d of z is sai d to be d iff at z iff lim
f ( z h ) f ( z )
h
exists uniq uely ,
w hen h thr ou g h co m p lex v alues. Th is li m is called d eriv at ive of f at z i s d en ote d b y f ( z )
? f ( z ) lim
f ( z h ) f ( z )
h
Thm 3.1 Let f = u+iv be diff at z. Then f satisfy C-R eq
n
f y =i f x & converse is not true.
Proof: Let the fun
c
f is diff at z.
lim
f ( z h ) f ( z )
h
exist uniq uely as h along any d ir
Let z = x + i y & h=h 1+i h 2
ake h along r ea l axis, then h 2= 0
? f ( z ) lim
f ( x h
h , y ) f ( x , y )
h h
= f x … … … … … … … … … … … (i)
ake h along imag in ary axis then h 1
? f
( z ) lim
f ( x , y h ) f ( x , y )
h
lim
f ( x , y h ) f ( x , y )
i h
i
lim
f ( x , y h ) f ( x , y )
h
i
f
… … … ( ii )
? f r om ( i) ( ii )
i f x = f y
but converse of the above thm is not true.
i.e. f x & f y may exists in a nbhd of z & may satisfy C-R eq
n
f y=i f x but the fun
c
f is not diff at z.
E xample f ( x ) {
xy ( x iy )
x
y
, if z , if z
Now C-R eq
n
at z=0
f
lim
f ( h , ) f ( , )
h
lim
h
lim
h
(
* f or m
lim
by L ’Hop it a l r ule
f
lim
f ( , h ) f ( , )
h
lim
f ( , h ) f ( , )
h
lim
h
lim
h
(
* in d et erm in at e f or m
lim
H en ce C R eq n sat isfie d b ut not d iff at z
Consider,
lim
f ( z ) f ( )
z lim
xy ( x iy )
x
y
x i y
lim
xy
x
y
lim
( , ) ( , )
xy
x
y
Taking along the path y=mx as x , y
lim
( , ) ( , )
x mx
x
( mx )
lim
( , ) ( , )
x mx
x
( m
)
h =h 1+0 i
h real ax i s t h en h 1
i f h al ong im ag i n ary a x i s t he n h 2
also h= 0+ih 2
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lim
m
m
m
m
f or d iff v alue of m we f in d d iff erent lim
lim is not uniq ue .
? f is n ot d iff at z .
Note (i) f is said to be analytic at z iff f is diff in the nbhd of z.
(ii) f is said to be analytic in a set S if f is diff in an open set containing S.
3.6 Proposition
If f = u+iv be analytic in a region D and u is constant, and then f is constant.
Proof let f be analytic in a region D.
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+iv x)
v y = u x
u y v x …… ( i)
given: u consta nt
u x = 0, u y =0
v y = 0, v x =0 by (i)
v consta nt
f is const an t.
Result let f = u+i v be analytic in a region D. Let v is constant, and then f is constant.
Proof let f be analytic in a region D
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+i v x)
u y v x
v y = u x
now, It is given that v is constant
p artia l d eriv at ive of v mu st b e zer o
v x=v y =0
u x = u y =0 b y C R eq
n
i.e. partial derivative of u(x, y) are zero
u is consta nt hen c e f is const an t
3.7 Proposition
If f =u+i v be analytic in a region D and if |f| be constant there, then f is constant.
Sol: let f = u+iv
? |f| = v u
v
= constant
u
2
+v
2
c ( say) …… …… . . ( i)
Case 1 if c = 0
? u
2
+v
2
= 0
u
2
= 0 & v
2
= 0
u = 0 & v = 0
u i v = 0
f = 0 f is const an t
Case 2 if c
Diff partially of eq
n
(i) w.r . t. ‘x ’ ‘y ’
2 u u x + 2 v v x = 0
& 2 u u y + 2 v v y = 0
u u x + v v x = 0
& u u y + v v y = 0
u v y + v v x …… . . ( ii )
u v x + v v y …… …. . ( i ii )
Multiply eq
n
(ii) by v, eq
n
(iii) by u and add
u v y +v v x = 0
u ( v x) + v v y =0
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(u
2
+v
2
)v y = 0
v y ( u
2
+v
2
)
Similarly, u y = 0
? b y C R eq
n
u x v x = 0
? u x = u y = v x = v y = 0
u consta nt v co nst an t
f is const an t
LMV If a fun
c
f is defined on a [a, a+h] s.t.
(i) f continuous on [a, a+h]
(ii) f d iff ( a, a h) the n ? ( , ) s. t.
f(a+h) – f(a) = h f
( a h)
for 2 variable
f(a+h, b) – f(a, b) = h f x ( a h , b )
f(a, b+h) – f(a, b) = h f y ( a, b h)
Thm 3.2 Let f x & f y exists in a nbhd of z, Let f x & f y b e ct s at z , le t f sat isfy C R eq
n
f y = i f x
i.e. u x = v y
u y v x
proof let f = u+i v , z = x +i y
we have to prove
lim
f ( z h ) f ( z )
h
f
( z )
L et h i z = x+i y
z h x i ( y )
consider,
f ( z h ) f ( z )
h
u ( x , y ) i v ( x , y ) - ,u ( x , y ) i v ( x , y ) -
i
u ( x , y ) u ( x , y ) - i , v ( x , y ) v ( x , y ) -
i … . ( i )
Now,
u( x , y ) u( x, y) u( x , y ) u ( x , y ) u ( x , y )
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
–u(x, y)
,u( x , y ) u( x, y )- ,u( x, y ) u ( x, y ) -
u x ( x 1 , y ) u y(x, y 2 )
w her e 1 , 2 by L V …… ( ii)
u x(z 1 ) u y(z 2)
Where z 1 ( x 1 ) i (y )
z 2 x i ( y 2 )
&|z 1 z | | z 2 z | as h or ,
Similarly,
v ( x , y ) v ( x, y) v x(z 3 ) v y(z 4) …… …. . ( iii)
where |z 3 z | | z 4 z | as h
from (i) , (ii) & (iii)
f ( z h ) f ( z )
h
u
( z
) u
( z
) - i , v
( z
) v
( z )
i
, u
( z
) i v
( z
) - ,u
( z
) i v
( z
)
i … ( iv )
Now, f x(z) = u x(z)+i v x(z)
,u x ( z ) i v x ( z ) - i i
,u
( z ) i v
( z ) - , i u
( z ) v
( z ) -
i
Then f is diff at z & f
(z) = f x(z)
suppose f is analytic in a region R, if one of
the foll
n
hold
(1) Im f is constant on
(2) Re f is constant on
(3) |f| is constant on
Then f is constant on
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,u
( z ) i v
( z ) - , u
( z ) i v
( z ) -
i by C R eq uati on … . ( v )
f ( z h ) f ( z )
h
f
( z )
i , u
( z
) i v
( z
) * u
( z ) i v
( z ) + - i [ u
( z
) i v
( z
) { u
( z ) i v
( z ) } ]
|
f ( z h ) f ( z )
h
f
( z ) | |
i | ,| u
( z
) u
( z ) | | v
( z
) v
( z ) | -
|
i | [ | u
( z
) u
( z ) | | v
( z
) v
( z ) | ]
as h
{
|
| , |
| u
, u
, v
, v
are cts at z b eca use of f
, f
are cts at z
}
| z
z | | z
z | | z
z | | z
z | as h
Hence, f is diff at z (by def
n
)
f ( z ) f x(z)
Thm 3.5 Let g is the inverse of f at z 0, Let g be cts at z 0 . Let the fun
c
f is diff at g(z 0 ) f ( g ( z 0 ) ) . Th en g is
d iff at z
g ( z
) f
( g ( z
) )
Pro of consid er , g ( z ) g ( z
)
z z
g ( z ) g ( z
)
f ( g ( z ) ) f ( g ( z
)
… … … … … … ( i )
g is t he in v erse of f
ie g= f
-1
g (z) = f
-1
( z ) f ( g ( z ) ) z +
Let g (z 0) = w 0 , g(z)= w
the fu n
c
g is cts at z 0
? lim
g ( z ) g ( z
) lim
w w
ie w w 0 as z z 0
? eq uati on ( i ) lim
g ( z ) g ( z
)
z z
lim
w w
f ( w ) f ( w
)
lim
f ( w ) f ( w
)
w w
… … … … … … ( ii )
f is diff at g (z 0)=w 0
? y the def
n
of differentiability
lim
f ( w ) f ( w
)
w w
f ( w
)
? f r om eq
n
(i) & (ii)
lim
g ( z ) g ( z
)
z z
f
( w
)
f
( g ( z
) )
g is d iff at z
g ( z
) f
( g ( z
) )
Ex f(z)=|z|
2
Let z = x+i y
? f ( z ) | z |
2
= |x +i y|
2
=x
2
+y
2
? u(x, y)=x
2
+y
2
, v ( x, y ) ( f ( x) u iv
? u x =2 x , u y=2 y
v x =0, v y=0
if ( x, y) ( , )
then C-R eq
n
are not sat isfied f or z
? the fu n
n
is n ot d iff f o r z
Page 5
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CHAPTER 3 ANALYTIC FUNCTION
Defini ti on f un c d efine d in a ne i g hbor hoo d of z is sai d to be d iff at z iff lim
f ( z h ) f ( z )
h
exists uniq uely ,
w hen h thr ou g h co m p lex v alues. Th is li m is called d eriv at ive of f at z i s d en ote d b y f ( z )
? f ( z ) lim
f ( z h ) f ( z )
h
Thm 3.1 Let f = u+iv be diff at z. Then f satisfy C-R eq
n
f y =i f x & converse is not true.
Proof: Let the fun
c
f is diff at z.
lim
f ( z h ) f ( z )
h
exist uniq uely as h along any d ir
Let z = x + i y & h=h 1+i h 2
ake h along r ea l axis, then h 2= 0
? f ( z ) lim
f ( x h
h , y ) f ( x , y )
h h
= f x … … … … … … … … … … … (i)
ake h along imag in ary axis then h 1
? f
( z ) lim
f ( x , y h ) f ( x , y )
h
lim
f ( x , y h ) f ( x , y )
i h
i
lim
f ( x , y h ) f ( x , y )
h
i
f
… … … ( ii )
? f r om ( i) ( ii )
i f x = f y
but converse of the above thm is not true.
i.e. f x & f y may exists in a nbhd of z & may satisfy C-R eq
n
f y=i f x but the fun
c
f is not diff at z.
E xample f ( x ) {
xy ( x iy )
x
y
, if z , if z
Now C-R eq
n
at z=0
f
lim
f ( h , ) f ( , )
h
lim
h
lim
h
(
* f or m
lim
by L ’Hop it a l r ule
f
lim
f ( , h ) f ( , )
h
lim
f ( , h ) f ( , )
h
lim
h
lim
h
(
* in d et erm in at e f or m
lim
H en ce C R eq n sat isfie d b ut not d iff at z
Consider,
lim
f ( z ) f ( )
z lim
xy ( x iy )
x
y
x i y
lim
xy
x
y
lim
( , ) ( , )
xy
x
y
Taking along the path y=mx as x , y
lim
( , ) ( , )
x mx
x
( mx )
lim
( , ) ( , )
x mx
x
( m
)
h =h 1+0 i
h real ax i s t h en h 1
i f h al ong im ag i n ary a x i s t he n h 2
also h= 0+ih 2
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lim
m
m
m
m
f or d iff v alue of m we f in d d iff erent lim
lim is not uniq ue .
? f is n ot d iff at z .
Note (i) f is said to be analytic at z iff f is diff in the nbhd of z.
(ii) f is said to be analytic in a set S if f is diff in an open set containing S.
3.6 Proposition
If f = u+iv be analytic in a region D and u is constant, and then f is constant.
Proof let f be analytic in a region D.
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+iv x)
v y = u x
u y v x …… ( i)
given: u consta nt
u x = 0, u y =0
v y = 0, v x =0 by (i)
v consta nt
f is const an t.
Result let f = u+i v be analytic in a region D. Let v is constant, and then f is constant.
Proof let f be analytic in a region D
f is d iff in a se t S D
ever y d iff f un
c
satisfy C-R eq
n
f y = i f x
u y+i v y = i (u x+i v x)
u y v x
v y = u x
now, It is given that v is constant
p artia l d eriv at ive of v mu st b e zer o
v x=v y =0
u x = u y =0 b y C R eq
n
i.e. partial derivative of u(x, y) are zero
u is consta nt hen c e f is const an t
3.7 Proposition
If f =u+i v be analytic in a region D and if |f| be constant there, then f is constant.
Sol: let f = u+iv
? |f| = v u
v
= constant
u
2
+v
2
c ( say) …… …… . . ( i)
Case 1 if c = 0
? u
2
+v
2
= 0
u
2
= 0 & v
2
= 0
u = 0 & v = 0
u i v = 0
f = 0 f is const an t
Case 2 if c
Diff partially of eq
n
(i) w.r . t. ‘x ’ ‘y ’
2 u u x + 2 v v x = 0
& 2 u u y + 2 v v y = 0
u u x + v v x = 0
& u u y + v v y = 0
u v y + v v x …… . . ( ii )
u v x + v v y …… …. . ( i ii )
Multiply eq
n
(ii) by v, eq
n
(iii) by u and add
u v y +v v x = 0
u ( v x) + v v y =0
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(u
2
+v
2
)v y = 0
v y ( u
2
+v
2
)
Similarly, u y = 0
? b y C R eq
n
u x v x = 0
? u x = u y = v x = v y = 0
u consta nt v co nst an t
f is const an t
LMV If a fun
c
f is defined on a [a, a+h] s.t.
(i) f continuous on [a, a+h]
(ii) f d iff ( a, a h) the n ? ( , ) s. t.
f(a+h) – f(a) = h f
( a h)
for 2 variable
f(a+h, b) – f(a, b) = h f x ( a h , b )
f(a, b+h) – f(a, b) = h f y ( a, b h)
Thm 3.2 Let f x & f y exists in a nbhd of z, Let f x & f y b e ct s at z , le t f sat isfy C R eq
n
f y = i f x
i.e. u x = v y
u y v x
proof let f = u+i v , z = x +i y
we have to prove
lim
f ( z h ) f ( z )
h
f
( z )
L et h i z = x+i y
z h x i ( y )
consider,
f ( z h ) f ( z )
h
u ( x , y ) i v ( x , y ) - ,u ( x , y ) i v ( x , y ) -
i
u ( x , y ) u ( x , y ) - i , v ( x , y ) v ( x , y ) -
i … . ( i )
Now,
u( x , y ) u( x, y) u( x , y ) u ( x , y ) u ( x , y )
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯
–u(x, y)
,u( x , y ) u( x, y )- ,u( x, y ) u ( x, y ) -
u x ( x 1 , y ) u y(x, y 2 )
w her e 1 , 2 by L V …… ( ii)
u x(z 1 ) u y(z 2)
Where z 1 ( x 1 ) i (y )
z 2 x i ( y 2 )
&|z 1 z | | z 2 z | as h or ,
Similarly,
v ( x , y ) v ( x, y) v x(z 3 ) v y(z 4) …… …. . ( iii)
where |z 3 z | | z 4 z | as h
from (i) , (ii) & (iii)
f ( z h ) f ( z )
h
u
( z
) u
( z
) - i , v
( z
) v
( z )
i
, u
( z
) i v
( z
) - ,u
( z
) i v
( z
)
i … ( iv )
Now, f x(z) = u x(z)+i v x(z)
,u x ( z ) i v x ( z ) - i i
,u
( z ) i v
( z ) - , i u
( z ) v
( z ) -
i
Then f is diff at z & f
(z) = f x(z)
suppose f is analytic in a region R, if one of
the foll
n
hold
(1) Im f is constant on
(2) Re f is constant on
(3) |f| is constant on
Then f is constant on
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,u
( z ) i v
( z ) - , u
( z ) i v
( z ) -
i by C R eq uati on … . ( v )
f ( z h ) f ( z )
h
f
( z )
i , u
( z
) i v
( z
) * u
( z ) i v
( z ) + - i [ u
( z
) i v
( z
) { u
( z ) i v
( z ) } ]
|
f ( z h ) f ( z )
h
f
( z ) | |
i | ,| u
( z
) u
( z ) | | v
( z
) v
( z ) | -
|
i | [ | u
( z
) u
( z ) | | v
( z
) v
( z ) | ]
as h
{
|
| , |
| u
, u
, v
, v
are cts at z b eca use of f
, f
are cts at z
}
| z
z | | z
z | | z
z | | z
z | as h
Hence, f is diff at z (by def
n
)
f ( z ) f x(z)
Thm 3.5 Let g is the inverse of f at z 0, Let g be cts at z 0 . Let the fun
c
f is diff at g(z 0 ) f ( g ( z 0 ) ) . Th en g is
d iff at z
g ( z
) f
( g ( z
) )
Pro of consid er , g ( z ) g ( z
)
z z
g ( z ) g ( z
)
f ( g ( z ) ) f ( g ( z
)
… … … … … … ( i )
g is t he in v erse of f
ie g= f
-1
g (z) = f
-1
( z ) f ( g ( z ) ) z +
Let g (z 0) = w 0 , g(z)= w
the fu n
c
g is cts at z 0
? lim
g ( z ) g ( z
) lim
w w
ie w w 0 as z z 0
? eq uati on ( i ) lim
g ( z ) g ( z
)
z z
lim
w w
f ( w ) f ( w
)
lim
f ( w ) f ( w
)
w w
… … … … … … ( ii )
f is diff at g (z 0)=w 0
? y the def
n
of differentiability
lim
f ( w ) f ( w
)
w w
f ( w
)
? f r om eq
n
(i) & (ii)
lim
g ( z ) g ( z
)
z z
f
( w
)
f
( g ( z
) )
g is d iff at z
g ( z
) f
( g ( z
) )
Ex f(z)=|z|
2
Let z = x+i y
? f ( z ) | z |
2
= |x +i y|
2
=x
2
+y
2
? u(x, y)=x
2
+y
2
, v ( x, y ) ( f ( x) u iv
? u x =2 x , u y=2 y
v x =0, v y=0
if ( x, y) ( , )
then C-R eq
n
are not sat isfied f or z
? the fu n
n
is n ot d iff f o r z
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If (x, y)=(0, 0)
Then C-R eq
n
satisfied at z=0
? f un
c
is diff at z=0 ( u (x, y) & v(x, y) both are cts)
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