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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
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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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? ? 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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FAQs on Line Integral - Topic-wise Tests & Solved Examples for Mathematics
| 1. What is a line integral in mathematics? |  |
A line integral is a mathematical concept that calculates the cumulative effect of a vector field along a given curve. It measures the total work done or the total accumulated value of a quantity as one moves along the curve.
| 2. How is a line integral calculated? | |
To calculate a line integral, one needs to divide the given curve into small segments and approximate the integral over each segment. The approximation is improved by taking smaller segments. The line integral is then obtained by summing up the approximations for all segments.
| 3. What is the significance of line integrals in physics? | |
Line integrals play a crucial role in physics, particularly in the field of electromagnetism. They are used to calculate quantities such as work done by a force along a path, electric or magnetic field strength, and circulation of a vector field. Line integrals are essential in understanding the behavior of fields and forces in physical systems.
| 4. Can line integrals be negative? | |
Yes, line integrals can be negative. The value of a line integral depends on the orientation of the curve and the vector field being integrated. If the curve and the vector field have opposite directions, the line integral can result in a negative value. It signifies that the work done or the accumulated quantity is in the opposite direction of the curve.
| 5. What are some real-life applications of line integrals? | |
Line integrals find applications in various real-life scenarios. They are used in physics to calculate the work done by a force in moving an object along a curved path. In engineering, line integrals are used to analyze fluid flow, such as calculating the circulation of a fluid in a closed loop. They also have applications in computer graphics, where they are used to render three-dimensional objects by simulating the movement of light along curves.
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