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# Line Integral Mathematics Notes | EduRev

## Mathematics : Line Integral Mathematics Notes | EduRev

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? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t   it ) dt
? ( t
i t
i t
t
) dt

i ? t
dt
i 4
t
5
i
i
3

Ex.1\Pg.46       Evaluate     , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t        t
Sol
n
:- z ( t)  t it          z  ( t )     i
But   z = x+iy
x =t , y=t
? f ( z )   ( x(t ) )
2
+i (y(t))
2

=t
2
+ i t
2

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) (  i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE:  Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z )  z
C is g iven by z ( t ) sin t i cos t ,  t
S oltio n  f ( z )   z
x iy  x i y
x
y
x
x
y
y
x
y

z(t) = sin t + i cost   z  ( t)   cos t – i sin t
but z = x +i y
? x   sin t ,   y = cos t
?  x
2
+ y
2
=1
f(z(t))  = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt

? ( sin t cost i ) dt

Ex   pg    E v aluate ? f
w her e f ( z )  z
C is g iven by z ( t ) R cos t i R sin t ,  t   , R
S olu ti on f ( z )  z
x i y
x i y x
y

z ( t ) R cos t i R sin t z
( t )  R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos  i sin R

( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt

? ( sin t cos t i sin
t i cos
t cos t sin t ) dt

? ( sin
t cos
t ) dt

i ? . dt

i ( t )

i
T he cu r v e i s sai d to b e smooth iff z ( t)   , excep t a t a f in it e n o. of p oi nt s.
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? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t   it ) dt
? ( t
i t
i t
t
) dt

i ? t
dt
i 4
t
5
i
i
3

Ex.1\Pg.46       Evaluate     , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t        t
Sol
n
:- z ( t)  t it          z  ( t )     i
But   z = x+iy
x =t , y=t
? f ( z )   ( x(t ) )
2
+i (y(t))
2

=t
2
+ i t
2

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) (  i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE:  Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z )  z
C is g iven by z ( t ) sin t i cos t ,  t
S oltio n  f ( z )   z
x iy  x i y
x
y
x
x
y
y
x
y

z(t) = sin t + i cost   z  ( t)   cos t – i sin t
but z = x +i y
? x   sin t ,   y = cos t
?  x
2
+ y
2
=1
f(z(t))  = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt

? ( sin t cost i ) dt

Ex   pg    E v aluate ? f
w her e f ( z )  z
C is g iven by z ( t ) R cos t i R sin t ,  t   , R
S olu ti on f ( z )  z
x i y
x i y x
y

z ( t ) R cos t i R sin t z
( t )  R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos  i sin R

( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt

? ( sin t cos t i sin
t i cos
t cos t sin t ) dt

? ( sin
t cos
t ) dt

i ? . dt

i ( t )

i
T he cu r v e i s sai d to b e smooth iff z ( t)   , excep t a t a f in it e n o. of p oi nt s.
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i. e.              iff x ( t) y ( t)   ,
Smoothly equivalent Curve Curve C 1 g iven b y z   z ( t)   a  t   b is s . t . b . smo ot hly equivalent to Curve C 2 given
b y w   w ( t)   c   t   d . If there exist a on e t o one cont in uou sly d iff ma p p in g
?    ,c , d -  ,a , b - s. t .
(i) ? ( c)   a ,    ? ( d ) b ,  ? ( t ) =   ?  t   increasing
(ii) z ( ?( t) )   w ( t )
Que. 1 Prove that a smoothly equivalent curve is an equivalence relation.
Reflexive C given by z = z(t) is smoothly equivalent to itself.
S in ce ? can b e t aken a s i d en ti ty f un
c
.
Symmetric let C 1 be smoothly equivalent to C 2.
Let  ?   ,c, d -  ,a , b -  b e one to one sat isfy in g cod
n

?( c) a  ?( d ) b & z ( ?( t ) ) w ( t)  then
?
1
,a, b -  ,c, d - w ill g iv es C 2 which is also smoothly equivalent to C 1
Transitive  let C 1 : z=z(t) ,   a   t   b
C 2   w w ( t)  c   t   d
C 3   r   r ( t)    e   t   f
If C 1 be smoothly equivalent to C 2
? ? a one to one map p i ng ? 1  ,c, d -  ,a , b -  s. t .
? 1 ( c)  a,   ? 1 ( d ) b    z ( ? 1(t))=w(t)
and C 2 be smoothly equivalent to C 3 then
? 2 ( e ) c , ? 2 ( f ) d      w ( ? 2(t))=r(t)
let ? ? 1 ° ? 2 then ? i s on e t o one such that
?   ,e, f -  ,a , b -
?( e) ( ? 1 ° ? 2 ) ( e) ? 1 (? 2 ( e) ) ? 1(c)=a
?( f )  ( ? 1 ° ? 2 ) ( f )  ? 1 (? 2 ( f ) ) ? 1(d)=b
also,  z ( ?( t) ) z ( ( ? 1 ° ? 2)(t))
z ( ? 1 (? 2 ( t) ) )   w ( ? 2(t)) = r(t)
Therefore, C 1 is smoothly equivalent to C 3
Hence, a smoothly equivalent curve is an equivalence relation.
T hm  .  If C
C
are smo othly eq uiv ale nt cu r v e , then ? f
c
? f
c

Proof let C 1 :z = z(t) , a   t   b
C 2 :w = w(t) ,  c   t   d
Let C 1 & C 2 be smoothly equivalent, therefore there exists a mapping
?   ,c, d -  ,a , b - s. t .
?( c)  a ,  ?( d ) b       z ( ?( t ) ) w ( t)
Consider,
? ? f
? f ( z ) dz
? f ( z ( t ) ) z
( t ) dt

= , u
( z ( t) ) iv ( z ( t) ) -,x ( t ) iy ( t) -d t
= , u
( z ( t) ) x ( t ) d t  v
( z ( t) ) y ( t) d t-  i, u
( z ( t) ) y ( t ) d t   v
( z ( t) ) x ( t ) d t …… …( i)
Consider,
? ? f
? f ( w ) dw
? f ( w ( t ) ) w
( t ) dt

= f
( z ( ? ( t) ) ) z (?( t ) ) ? ( t ) )
= , u
( z ( ? ( t) ) ) i v( z ( ?(t) ) ) - ,x ( ?( t ) ) i y ( ? ( t) ) -? ( t) d t
=[ u
( z ( ? ( t) ) x ( ? ( t) ) ? ( t) d t   v
( z ( ? ( t) ) ) y ( ?( t ) ) ? ( t ) d t-   i, u
( z ( ?( t) ) ) y ( ?(t) ) ? ( t) d t
+  v
( z ( ?(t) ) ) x ( ?( t ) ) ? ( t ) d t-   ……( ii )
eq
n
(i) & (ii) are equal
by change of variable of real integral
ls o , ? x ( t )
dt ? y ( t )
????                    y c ha n ge of va r ia b l e
Put ?(t) = r When
t =c    then  ?(c)=a & t =d    then ?(d)=b
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? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t   it ) dt
? ( t
i t
i t
t
) dt

i ? t
dt
i 4
t
5
i
i
3

Ex.1\Pg.46       Evaluate     , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t        t
Sol
n
:- z ( t)  t it          z  ( t )     i
But   z = x+iy
x =t , y=t
? f ( z )   ( x(t ) )
2
+i (y(t))
2

=t
2
+ i t
2

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) (  i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE:  Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z )  z
C is g iven by z ( t ) sin t i cos t ,  t
S oltio n  f ( z )   z
x iy  x i y
x
y
x
x
y
y
x
y

z(t) = sin t + i cost   z  ( t)   cos t – i sin t
but z = x +i y
? x   sin t ,   y = cos t
?  x
2
+ y
2
=1
f(z(t))  = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt

? ( sin t cost i ) dt

Ex   pg    E v aluate ? f
w her e f ( z )  z
C is g iven by z ( t ) R cos t i R sin t ,  t   , R
S olu ti on f ( z )  z
x i y
x i y x
y

z ( t ) R cos t i R sin t z
( t )  R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos  i sin R

( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt

? ( sin t cos t i sin
t i cos
t cos t sin t ) dt

? ( sin
t cos
t ) dt

i ? . dt

i ( t )

i
T he cu r v e i s sai d to b e smooth iff z ( t)   , excep t a t a f in it e n o. of p oi nt s.
Free coaching of B.Sc (h) maths & JAM
For more 8130648819

i. e.              iff x ( t) y ( t)   ,
Smoothly equivalent Curve Curve C 1 g iven b y z   z ( t)   a  t   b is s . t . b . smo ot hly equivalent to Curve C 2 given
b y w   w ( t)   c   t   d . If there exist a on e t o one cont in uou sly d iff ma p p in g
?    ,c , d -  ,a , b - s. t .
(i) ? ( c)   a ,    ? ( d ) b ,  ? ( t ) =   ?  t   increasing
(ii) z ( ?( t) )   w ( t )
Que. 1 Prove that a smoothly equivalent curve is an equivalence relation.
Reflexive C given by z = z(t) is smoothly equivalent to itself.
S in ce ? can b e t aken a s i d en ti ty f un
c
.
Symmetric let C 1 be smoothly equivalent to C 2.
Let  ?   ,c, d -  ,a , b -  b e one to one sat isfy in g cod
n

?( c) a  ?( d ) b & z ( ?( t ) ) w ( t)  then
?
1
,a, b -  ,c, d - w ill g iv es C 2 which is also smoothly equivalent to C 1
Transitive  let C 1 : z=z(t) ,   a   t   b
C 2   w w ( t)  c   t   d
C 3   r   r ( t)    e   t   f
If C 1 be smoothly equivalent to C 2
? ? a one to one map p i ng ? 1  ,c, d -  ,a , b -  s. t .
? 1 ( c)  a,   ? 1 ( d ) b    z ( ? 1(t))=w(t)
and C 2 be smoothly equivalent to C 3 then
? 2 ( e ) c , ? 2 ( f ) d      w ( ? 2(t))=r(t)
let ? ? 1 ° ? 2 then ? i s on e t o one such that
?   ,e, f -  ,a , b -
?( e) ( ? 1 ° ? 2 ) ( e) ? 1 (? 2 ( e) ) ? 1(c)=a
?( f )  ( ? 1 ° ? 2 ) ( f )  ? 1 (? 2 ( f ) ) ? 1(d)=b
also,  z ( ?( t) ) z ( ( ? 1 ° ? 2)(t))
z ( ? 1 (? 2 ( t) ) )   w ( ? 2(t)) = r(t)
Therefore, C 1 is smoothly equivalent to C 3
Hence, a smoothly equivalent curve is an equivalence relation.
T hm  .  If C
C
are smo othly eq uiv ale nt cu r v e , then ? f
c
? f
c

Proof let C 1 :z = z(t) , a   t   b
C 2 :w = w(t) ,  c   t   d
Let C 1 & C 2 be smoothly equivalent, therefore there exists a mapping
?   ,c, d -  ,a , b - s. t .
?( c)  a ,  ?( d ) b       z ( ?( t ) ) w ( t)
Consider,
? ? f
? f ( z ) dz
? f ( z ( t ) ) z
( t ) dt

= , u
( z ( t) ) iv ( z ( t) ) -,x ( t ) iy ( t) -d t
= , u
( z ( t) ) x ( t ) d t  v
( z ( t) ) y ( t) d t-  i, u
( z ( t) ) y ( t ) d t   v
( z ( t) ) x ( t ) d t …… …( i)
Consider,
? ? f
? f ( w ) dw
? f ( w ( t ) ) w
( t ) dt

= f
( z ( ? ( t) ) ) z (?( t ) ) ? ( t ) )
= , u
( z ( ? ( t) ) ) i v( z ( ?(t) ) ) - ,x ( ?( t ) ) i y ( ? ( t) ) -? ( t) d t
=[ u
( z ( ? ( t) ) x ( ? ( t) ) ? ( t) d t   v
( z ( ? ( t) ) ) y ( ?( t ) ) ? ( t ) d t-   i, u
( z ( ?( t) ) ) y ( ?(t) ) ? ( t) d t
+  v
( z ( ?(t) ) ) x ( ?( t ) ) ? ( t ) d t-   ……( ii )
eq
n
(i) & (ii) are equal
by change of variable of real integral
ls o , ? x ( t )
dt ? y ( t )
????                    y c ha n ge of va r ia b l e
Put ?(t) = r When
t =c    then  ?(c)=a & t =d    then ?(d)=b
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?     f
=    f

ALTERNATE
Let C 1 : z = z(t)  , a   t   b
C 2 : w= w(t)  , c   t   d
Suppose C 1 & C 2 be smoothly equivalent Curve,
? ?   ,c, d -  ,a , b -  s. t.
?( c)  = a , ?( d ) b  &   z ( ?( t) w ( t)
Consider,
c
f =  f
(w)dw =  f
( w ( t) ) w ( t) d t
= f
( z ( ? ( t) ) ) z (?( t ) ) ) ? ( t) d t   p ut ?( t )   p
= f
( z ( p ) ) z (p) d p       ? ( t) d t  d p
c
f (z)dz
Def
n
.  let C b e t he cu r v e giv en b y z  z ( t)   a   t   b then
C  is d efine d b y z ( b a t)   a   t   b .
(intuitively ?????????? ?????? , ) C has the sa me p o in ts se t of C , traced in th e rev erse d ir
n
.
ote   C ler aly , C ( b )    C ( a)
C( a)    C ( b )
F ur ther, ( c) ( t )  C ( b a t )
T heor em  .   Pro v e that ? f
? f

Proof : Consider,
c f      f
( z ( b a t ) ) z (b a t) d t
, , * u
( z ( b a t ) )
i v( z ( b a t ) ) +* x ( b a t) ) iy ( b a t ) ) +--dt
, , u
( z ( b a t ) ) x ( b a t)
i v( z ( b a t ) ) x ( b a t ) ) i u ( b a t) y ( b a t) v ( z ( b a t ) y ( b a t) --d t
Put b a t   w   limit when  t = a ,  w = b
d t    d w     t = b , w  = a
dw   d t
? , u( z ( w ) x ( w ) i v ( z ( w ) ) x ( w ) i u ( z ( w ) ) y ( w ) v ( z ( w ) ) y ( w ) - dw

f
( z ( w ) ) z ( w ) d w     c f ( w ) d w     c f
Alternate
c f = f
( z ( b a t) ) z ( b a t) ( ) d t
Put b a t  w
d t  d w
f
( z ( t) ) z (t )dt   by the property of finite integral
=  -  c f   alternate on remaing portion paper
E x 3 /p g   S upp ose f( z )     let C b e smooth c ur v e , then  c f (z)dz =?
Sol
n
c f (z)dz=   f
(z)dz =  f
( z ) z ( t ) d t       z ( t) d t
= , z ( t ) -
z ( b )  z ( a)
NOTE let ?   ?? b e comp lex num b er then ? « ß if |? | | ?? |
Thm 4.9 Let G(t) be a continuous complex fun
n
of t, then
G
( t) d t «  | G |
dt       ( absolute value inequality ,  G
( t) d t    )
Proof: let    G
(t)dt =Re
i            R    …… . ( i)
Then,  e
G
(t)dt   =e
i  R e
i     R      … …. ( ii )
If  G
?? (t)dt =0, then above
inequality is obviously true. If not
we write its value in polar form, say
eq
n
(i)
? | G
?? (t)dt| = R
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? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) ( t   it ) dt
? ( t
i t
i t
t
) dt

i ? t
dt
i 4
t
5
i
i
3

Ex.1\Pg.46       Evaluate     , where f (z) =x
2
+iy
2
and C is given by
z ( t) t i t        t
Sol
n
:- z ( t)  t it          z  ( t )     i
But   z = x+iy
x =t , y=t
? f ( z )   ( x(t ) )
2
+i (y(t))
2

=t
2
+ i t
2

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( t
i t
) (  i ) dt
i ? t
dt
i 4
t
3
5
3
i
NOTE:  Curve in Que 2 & 3 both are smoothly equivalent by thm 4.5
Qu e 3 ? f
w her e f ( z )  z
C is g iven by z ( t ) sin t i cos t ,  t
S oltio n  f ( z )   z
x iy  x i y
x
y
x
x
y
y
x
y

z(t) = sin t + i cost   z  ( t)   cos t – i sin t
but z = x +i y
? x   sin t ,   y = cos t
?  x
2
+ y
2
=1
f(z(t))  = sin t –i cos t
? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
? ( sin t i cost ) ( cost i sin t ) dt

? ( sin t cost i ) dt

Ex   pg    E v aluate ? f
w her e f ( z )  z
C is g iven by z ( t ) R cos t i R sin t ,  t   , R
S olu ti on f ( z )  z
x i y
x i y x
y

z ( t ) R cos t i R sin t z
( t )  R sin t i R cos t
But, z =x+iy
? ( ( ) ) R cos R
R sin R
cos R
sin R

? ? f
? f ( z ) dz
? f ( z ( t ) )
z
( t ) dt
?
cos  i sin R

( R sin t i R cos t ) dt
? ( C ost iS in t ) ( S in t iCost ) dt

? ( sin t cos t i sin
t i cos
t cos t sin t ) dt

? ( sin
t cos
t ) dt

i ? . dt

i ( t )

i
T he cu r v e i s sai d to b e smooth iff z ( t)   , excep t a t a f in it e n o. of p oi nt s.
Free coaching of B.Sc (h) maths & JAM
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i. e.              iff x ( t) y ( t)   ,
Smoothly equivalent Curve Curve C 1 g iven b y z   z ( t)   a  t   b is s . t . b . smo ot hly equivalent to Curve C 2 given
b y w   w ( t)   c   t   d . If there exist a on e t o one cont in uou sly d iff ma p p in g
?    ,c , d -  ,a , b - s. t .
(i) ? ( c)   a ,    ? ( d ) b ,  ? ( t ) =   ?  t   increasing
(ii) z ( ?( t) )   w ( t )
Que. 1 Prove that a smoothly equivalent curve is an equivalence relation.
Reflexive C given by z = z(t) is smoothly equivalent to itself.
S in ce ? can b e t aken a s i d en ti ty f un
c
.
Symmetric let C 1 be smoothly equivalent to C 2.
Let  ?   ,c, d -  ,a , b -  b e one to one sat isfy in g cod
n

?( c) a  ?( d ) b & z ( ?( t ) ) w ( t)  then
?
1
,a, b -  ,c, d - w ill g iv es C 2 which is also smoothly equivalent to C 1
Transitive  let C 1 : z=z(t) ,   a   t   b
C 2   w w ( t)  c   t   d
C 3   r   r ( t)    e   t   f
If C 1 be smoothly equivalent to C 2
? ? a one to one map p i ng ? 1  ,c, d -  ,a , b -  s. t .
? 1 ( c)  a,   ? 1 ( d ) b    z ( ? 1(t))=w(t)
and C 2 be smoothly equivalent to C 3 then
? 2 ( e ) c , ? 2 ( f ) d      w ( ? 2(t))=r(t)
let ? ? 1 ° ? 2 then ? i s on e t o one such that
?   ,e, f -  ,a , b -
?( e) ( ? 1 ° ? 2 ) ( e) ? 1 (? 2 ( e) ) ? 1(c)=a
?( f )  ( ? 1 ° ? 2 ) ( f )  ? 1 (? 2 ( f ) ) ? 1(d)=b
also,  z ( ?( t) ) z ( ( ? 1 ° ? 2)(t))
z ( ? 1 (? 2 ( t) ) )   w ( ? 2(t)) = r(t)
Therefore, C 1 is smoothly equivalent to C 3
Hence, a smoothly equivalent curve is an equivalence relation.
T hm  .  If C
C
are smo othly eq uiv ale nt cu r v e , then ? f
c
? f
c

Proof let C 1 :z = z(t) , a   t   b
C 2 :w = w(t) ,  c   t   d
Let C 1 & C 2 be smoothly equivalent, therefore there exists a mapping
?   ,c, d -  ,a , b - s. t .
?( c)  a ,  ?( d ) b       z ( ?( t ) ) w ( t)
Consider,
? ? f
? f ( z ) dz
? f ( z ( t ) ) z
( t ) dt

= , u
( z ( t) ) iv ( z ( t) ) -,x ( t ) iy ( t) -d t
= , u
( z ( t) ) x ( t ) d t  v
( z ( t) ) y ( t) d t-  i, u
( z ( t) ) y ( t ) d t   v
( z ( t) ) x ( t ) d t …… …( i)
Consider,
? ? f
? f ( w ) dw
? f ( w ( t ) ) w
( t ) dt

= f
( z ( ? ( t) ) ) z (?( t ) ) ? ( t ) )
= , u
( z ( ? ( t) ) ) i v( z ( ?(t) ) ) - ,x ( ?( t ) ) i y ( ? ( t) ) -? ( t) d t
=[ u
( z ( ? ( t) ) x ( ? ( t) ) ? ( t) d t   v
( z ( ? ( t) ) ) y ( ?( t ) ) ? ( t ) d t-   i, u
( z ( ?( t) ) ) y ( ?(t) ) ? ( t) d t
+  v
( z ( ?(t) ) ) x ( ?( t ) ) ? ( t ) d t-   ……( ii )
eq
n
(i) & (ii) are equal
by change of variable of real integral
ls o , ? x ( t )
dt ? y ( t )
????                    y c ha n ge of va r ia b l e
Put ?(t) = r When
t =c    then  ?(c)=a & t =d    then ?(d)=b
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?     f
=    f

ALTERNATE
Let C 1 : z = z(t)  , a   t   b
C 2 : w= w(t)  , c   t   d
Suppose C 1 & C 2 be smoothly equivalent Curve,
? ?   ,c, d -  ,a , b -  s. t.
?( c)  = a , ?( d ) b  &   z ( ?( t) w ( t)
Consider,
c
f =  f
(w)dw =  f
( w ( t) ) w ( t) d t
= f
( z ( ? ( t) ) ) z (?( t ) ) ) ? ( t) d t   p ut ?( t )   p
= f
( z ( p ) ) z (p) d p       ? ( t) d t  d p
c
f (z)dz
Def
n
.  let C b e t he cu r v e giv en b y z  z ( t)   a   t   b then
C  is d efine d b y z ( b a t)   a   t   b .
(intuitively ?????????? ?????? , ) C has the sa me p o in ts se t of C , traced in th e rev erse d ir
n
.
ote   C ler aly , C ( b )    C ( a)
C( a)    C ( b )
F ur ther, ( c) ( t )  C ( b a t )
T heor em  .   Pro v e that ? f
? f

Proof : Consider,
c f      f
( z ( b a t ) ) z (b a t) d t
, , * u
( z ( b a t ) )
i v( z ( b a t ) ) +* x ( b a t) ) iy ( b a t ) ) +--dt
, , u
( z ( b a t ) ) x ( b a t)
i v( z ( b a t ) ) x ( b a t ) ) i u ( b a t) y ( b a t) v ( z ( b a t ) y ( b a t) --d t
Put b a t   w   limit when  t = a ,  w = b
d t    d w     t = b , w  = a
dw   d t
? , u( z ( w ) x ( w ) i v ( z ( w ) ) x ( w ) i u ( z ( w ) ) y ( w ) v ( z ( w ) ) y ( w ) - dw

f
( z ( w ) ) z ( w ) d w     c f ( w ) d w     c f
Alternate
c f = f
( z ( b a t) ) z ( b a t) ( ) d t
Put b a t  w
d t  d w
f
( z ( t) ) z (t )dt   by the property of finite integral
=  -  c f   alternate on remaing portion paper
E x 3 /p g   S upp ose f( z )     let C b e smooth c ur v e , then  c f (z)dz =?
Sol
n
c f (z)dz=   f
(z)dz =  f
( z ) z ( t ) d t       z ( t) d t
= , z ( t ) -
z ( b )  z ( a)
NOTE let ?   ?? b e comp lex num b er then ? « ß if |? | | ?? |
Thm 4.9 Let G(t) be a continuous complex fun
n
of t, then
G
( t) d t «  | G |
dt       ( absolute value inequality ,  G
( t) d t    )
Proof: let    G
(t)dt =Re
i            R    …… . ( i)
Then,  e
G
(t)dt   =e
i  R e
i     R      … …. ( ii )
If  G
?? (t)dt =0, then above
inequality is obviously true. If not
we write its value in polar form, say
eq
n
(i)
? | G
?? (t)dt| = R
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Let e
i
G(t) = A(t) + i B(t)
Where A(t) & B(t) are real value fun
n
of t.
e
G
(t)dt =     (t)+i     ( t) d t …. ( ii i )
? f r om eq
n
(ii) & (iii)
R=    (t)+i     (t)dt
(t)dt =R ,      (t)dt= 0
Now, R=     (t)dt =  r ea l ( e
G
(t))dt
| e
G
( t) | d t                   Re( z )   | z|
=  | e
| | G
(t)|dt
=  | G
(t)| dt
R       | G
( t) | d t     ……. (iv)
From eq
n
( i)   | G
(t)dt| = |Re
i  |
| G
(t)dt|  =R  …… …. . ( v )
?  from eq
n
(iv) & (v)
| G
(t)dt |<  | G |
dt
ALTERNATE
| G
(t)dt |<  | G |
dt
By polar form, also    G
(t)dt =R e
i
| G
(t)dt | =R  &   e
G
(t)dt=R
Taking the real part of both side of the above
R = Re (R) = Re( e
G
(t)dt)
= Re * e
G
(t)}dt
| e
G
(t)|dt  using R e( z )   | z |
= | G
(t)|dt     | e
i  | = 1
Hence, desired inequality hold.
.     L    F or mu la/   L   I ne q ulity  Suppose that C is a (smooth) curve of length L, that f    is continuous
on C, an d that ,  f      ( i. e . | f |   ) thr oug hou t C, then  c f ( z ) d z     L   i. e . | c f(z)dz| <ML     (M is the
upper bd of    f)
Proof: let the curve C be given by z = z(t) =x(t)+i y(t)
? | f
(z)dz| = | f
( z ( t) ) z (t) d t|
| f
( z ( t) ) z ( t) | d t
= | f
( z ( t) ) | | z ( t ) | d t     |
z ( t) | d t
=M |
x ( t) i y ( t) | d t     v *x ( t ) +
* y ( t )
+
dt
=M v *

+
*

+
dt =M

dt
=M ds
=ML    (where L = ds
= length of the Curve C or L=  v * x ( t ) +
* y ( t )
+
dt
? | f
( z ) d z |   L
Example/pg 48 let C be the unit circle and suppose f <<1 on C, then M=1, L=2  (length of the circle when
r  ) an d  c f(z)dz <<2  i. e. | c f ( z ) d z |
Sol
n
Here, M=1
L =length of the curve C = 2
y   L f or mu la
| c f ( z ) d z |    ? 2  =2
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## Topic-wise Tests & Solved Examples for IIT JAM Mathematics

27 docs|150 tests

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