Line Integral Mathematics Notes | EduRev

Topic-wise Tests & Solved Examples for IIT JAM Mathematics

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 
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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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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