Mathematics Exam > Mathematics Notes > Topic-wise Tests & Solved Examples for Mathematics > Analytic Function
Analytic Function | Topic-wise Tests & Solved Examples for Mathematics PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Page 3
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
(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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
(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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
,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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
(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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
,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
Free coaching of B.Sc (h) maths & JAM
For more 8130648819
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)
Read More
|
27 docs|150 tests
|
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.
Related Exams
About this Document
|
4.87/5 Rating |
|
Nov 21, 2024 Last updated |
Document Description: Analytic Function for Mathematics 2024 is part of Topic-wise Tests & Solved Examples for Mathematics preparation.
The notes and questions for Analytic Function have been prepared according to the Mathematics exam syllabus. Information about Analytic Function covers topics
like and Analytic Function Example, for Mathematics 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Analytic Function.
Introduction of Analytic Function in English is available as part of our Topic-wise Tests & Solved Examples for Mathematics
for Mathematics & Analytic Function in Hindi for Topic-wise Tests & Solved Examples for Mathematics course.
Download more important topics related with notes, lectures and mock test series for Mathematics
Exam by signing up for free. Mathematics: Analytic Function | Topic-wise Tests & Solved Examples for Mathematics
Description
Full syllabus notes, lecture & questions for Analytic Function | Topic-wise Tests & Solved Examples for Mathematics - Mathematics | Plus excerises question with solution to help you revise complete syllabus for Topic-wise Tests & Solved Examples for Mathematics | Best notes, free PDF download
Information about Analytic Function
In this doc you can find the meaning of Analytic Function defined & explained in the simplest way possible. Besides explaining types of
Analytic Function theory, EduRev gives you an ample number of questions to practice Analytic Function tests, examples and also practice Mathematics
tests
|
27 docs|150 tests
|
Download as PDF
|
Explore Courses for Mathematics exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
shortcuts and tricks
,Semester Notes
,Analytic Function | Topic-wise Tests & Solved Examples for Mathematics
,past year papers
,Analytic Function | Topic-wise Tests & Solved Examples for Mathematics
,Previous Year Questions with Solutions
,ppt
,Sample Paper
,Analytic Function | Topic-wise Tests & Solved Examples for Mathematics
,Free
,Extra Questions
,Objective type Questions
,study material
,video lectures
,Important questions
,Viva Questions
,MCQs
,mock tests for examination
,Exam
,Summary
,practice quizzes
;
Additional Information about Analytic Function for Mathematics Preparation
Analytic Function Free PDF Download
The Analytic Function is an invaluable resource that delves deep into the core of the Mathematics exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Analytic Function now and kickstart your journey towards success in the Mathematics exam.
Importance of Analytic Function
The importance of Analytic Function cannot be overstated, especially for Mathematics aspirants.
This document holds the key to success in the Mathematics exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Analytic Function Notes
Analytic Function Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Analytic Function.
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Analytic Function Notes on EduRev are your ultimate resource for success.
Analytic Function Mathematics Questions
The "Analytic Function Mathematics Questions" guide is a valuable resource for all aspiring students preparing for the
Mathematics exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Analytic Function on the App
Students of Mathematics can study Analytic Function alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Analytic Function,
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Analytic Function is prepared as per the latest Mathematics syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup