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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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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
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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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FAQs on Analytic Function - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is an analytic function in mathematics? |  |
Ans. An analytic function in mathematics refers to a function that can be locally represented by a convergent power series. It is a function that can be differentiated term by term within its interval of convergence.
| 2. How can I determine if a function is analytic? | |
Ans. To determine if a function is analytic, one can check for the Cauchy-Riemann equations. If a function satisfies these equations, it is analytic. Additionally, if a function can be represented by a power series that converges within a certain interval, it is also considered analytic within that interval.
| 3. What is the significance of analytic functions in complex analysis? | |
Ans. Analytic functions play a crucial role in complex analysis. They have the property that their derivative exists at every point in their domain, making them suitable for various mathematical applications. Analytic functions are used to solve complex equations, understand the behavior of complex numbers, and study the properties of complex functions.
| 4. Can an analytic function have singularities? | |
Ans. Yes, an analytic function can have singularities. These are points in the complex plane where the function is not defined or becomes infinite. Singularities can be classified into removable, poles, and essential singularities, depending on the behavior of the function near those points.
| 5. How are analytic functions related to harmonic functions? | |
Ans. Analytic functions and harmonic functions are closely related in complex analysis. If a function is analytic, it satisfies the Laplace's equation and is therefore harmonic. Conversely, if a function is harmonic and has continuous second-order partial derivatives, it can be locally represented as an analytic function. This connection allows for the interchangeability of techniques and results between the two areas of study.
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