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Exercise  
Q pg  3 / S how that  f
  lim
    ,      f ( z h ) f ( z )
h
, f
  lim
    ,     f ( z i h ) f ( z )
h
 
Provided the limit exist. 
Sol
n
:-   Let  us suppose that limit exist. 
 Let z=x+iy, also h be real  
 ? z h  ( x h,y) , or ( x h) iy  
 Z+ih =(x, y+h)  or x+i (y+h) 
?   lim
    ,      f ( x h ,   y ) f ( x ,   y )
h
 lim
    ,      f ( z h ) f ( z )
h
     … … … . . ( i ) 
     lim
    ,      f ( x ,   y h ) f ( x ,   y )
h
  lim
    ,      f ( z ih ) f ( z )
h
  … … … . . ( ii ) 
Hence  f x &  f y  are given by eq
n
 (i)& (ii) when the limit exist. 
Que. 2  a. Show that  f(z)=x
2
+iy
2 
 is diff at all points on the line y=x 
              b. Show that it is nowhere analytic. 
Sol
n
:- given f(z)=x
2
+iy
2 
 
?   f(x, y ) x
2
+iy
2
 
 ?   f x=2x & f y = i2y =2iy 
    f x & f y   are cts everywhere  
The fun
c
  f  is diff if it satisfy C-R eq
n 
 
ie.    f y= if x  
      2iy=(2x)i 
 Ie.    y=x 
Hence, f(z) is diff at all pts on the line y=x. 
Qu e.   S upp ose  f  is an a ly ti c in a reg ion   f      there. S how that  f is const an t. 
Sol
n
:-   Let   f=u+iv be analytic in a region 
 ? Dif f  w . r . t. ‘x ’   ‘y ’ p art ia lly . 
 ? f x =u x+iv x 
  f y= u y+iv y 
 g iven f     
   {
u
  iv
      u
  i v
              u x=0, v x=0, u y=0, v y=0. 
  all the p artia l d eriv at ive are eq ual to z ero 
 Hence,  f  is constant. 
Que. 6  Assume that fun
c
 f is analytic in a region & that at the every point of the  
r egion, ei ther f     or  f    . S how that f is const an t. ,H in t consi d er f
2
.] 
Sol
n
 :-     the fu n
n
  f is analytic 
 ? f
2
 is also analytic 
C onsider  ( f
 ( z ) )
    f ( z ) f
 ( z )        ( if  f ( z )    or f  ( z )    ) 
       (f
2
)          z 
 ?  y Prev ious que , f
2
 is constant & hence  f . 
Que 8. Find all analytic function f = u + iv with u(x, y) = x
2
-y
2
. 
So l
  Giv en u ( x , y ) x
  y
  
? u
   x 
  u
    y 
we k now that C auch y Rie mann eq uati on is  
u
  v
  
? v
   x 
in te g r at e p artia lly w . r . t . y 
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Exercise  
Q pg  3 / S how that  f
  lim
    ,      f ( z h ) f ( z )
h
, f
  lim
    ,     f ( z i h ) f ( z )
h
 
Provided the limit exist. 
Sol
n
:-   Let  us suppose that limit exist. 
 Let z=x+iy, also h be real  
 ? z h  ( x h,y) , or ( x h) iy  
 Z+ih =(x, y+h)  or x+i (y+h) 
?   lim
    ,      f ( x h ,   y ) f ( x ,   y )
h
 lim
    ,      f ( z h ) f ( z )
h
     … … … . . ( i ) 
     lim
    ,      f ( x ,   y h ) f ( x ,   y )
h
  lim
    ,      f ( z ih ) f ( z )
h
  … … … . . ( ii ) 
Hence  f x &  f y  are given by eq
n
 (i)& (ii) when the limit exist. 
Que. 2  a. Show that  f(z)=x
2
+iy
2 
 is diff at all points on the line y=x 
              b. Show that it is nowhere analytic. 
Sol
n
:- given f(z)=x
2
+iy
2 
 
?   f(x, y ) x
2
+iy
2
 
 ?   f x=2x & f y = i2y =2iy 
    f x & f y   are cts everywhere  
The fun
c
  f  is diff if it satisfy C-R eq
n 
 
ie.    f y= if x  
      2iy=(2x)i 
 Ie.    y=x 
Hence, f(z) is diff at all pts on the line y=x. 
Qu e.   S upp ose  f  is an a ly ti c in a reg ion   f      there. S how that  f is const an t. 
Sol
n
:-   Let   f=u+iv be analytic in a region 
 ? Dif f  w . r . t. ‘x ’   ‘y ’ p art ia lly . 
 ? f x =u x+iv x 
  f y= u y+iv y 
 g iven f     
   {
u
  iv
      u
  i v
              u x=0, v x=0, u y=0, v y=0. 
  all the p artia l d eriv at ive are eq ual to z ero 
 Hence,  f  is constant. 
Que. 6  Assume that fun
c
 f is analytic in a region & that at the every point of the  
r egion, ei ther f     or  f    . S how that f is const an t. ,H in t consi d er f
2
.] 
Sol
n
 :-     the fu n
n
  f is analytic 
 ? f
2
 is also analytic 
C onsider  ( f
 ( z ) )
    f ( z ) f
 ( z )        ( if  f ( z )    or f  ( z )    ) 
       (f
2
)          z 
 ?  y Prev ious que , f
2
 is constant & hence  f . 
Que 8. Find all analytic function f = u + iv with u(x, y) = x
2
-y
2
. 
So l
  Giv en u ( x , y ) x
  y
  
? u
   x 
  u
    y 
we k now that C auch y Rie mann eq uati on is  
u
  v
  
? v
   x 
in te g r at e p artia lly w . r . t . y 
Free coaching of B.Sc (h) maths & JAM 
For more 8130648819 
 
? v  xy c 
? f ( x
  y
 ) i ( xy c ) 
f ( x iy )
  ic 
f ( z ) z
  c
         ,    c
   . 
Qu e  . S how that there is no an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
So l
  L et if p ossi b le there is an an aly ti c f uncti on 
f u iv  w it h u ( x , y ) x
  y
  
if the f uncti on f is an aly ti c then it sat i sf ie s C R eq uati on . 
?
 u
 x
  x  v
 y
 v  xy c
             … … . . ( ) 
  u
 y
  y   v
 x
 v   xy c
     … … . . (  ) 
Which is a contradiction. 
? we can not f in d an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
 lte r na te 
also , ( v
 )
        ( v
 )
              usin g ( i )   ( ii )  
b ut v
  
 v
  
    sin ce f is an aly ti c  
?      a contrad ict ion . 
en ti r e  an aly C R e q
  
Qu e   . S upp ose f is an en t ire f uncti on of the f or m 
  f ( x , y ) u ( x ) iv ( y ) 
S how that f is a line ar p oly nomia l . 
So l
  L et f ( x , y ) u ( x ) iv ( y ) is an en t ire f uncti on .        … ( i ) 
? it is an aly ti c ov er w hol e d isc . 
? it sat isfy C auch y Rie mann eq uati on 
 ote en ti r e an aly ti c C R eq uati on sa ti sf ie d 
So , u
  v
  
i . e .  u
 ( x ) v
 ( y )    , u
 ( x ) is only f uncti on of x  v
 ( y ) is the only f uncti on of y - 
 u
 ( x ) and v
 ( y ) b oth are consta nt . 
L et u
 ( x ) v
 ( y ) a  
now , {
  ( )    ( )   
in te g r at e 
u ( x ) ax c
  
v ( y ) ay c
  
Put the value of u & v in equation (i)  
?  f ( x , y ) ( ax c
 ) i ( ay c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 az b      w her e  b c
  i c
  
Hence, f  is a linear polynomial. 
Qu e .  . a . S how that e
  is en ti r e by v erif y in g the C auch y Rie mann eq uati on f or its r ea l and imagina r y p arts . 
or  Pro v e that e
  is an a ly ti c in the case of r ea l and imagina r y p art . 
b . Pro v e  
e
      e
  e
   
So l
  L et f ( z ) e
  e
    
 e
 . e
    
u iv e
 ( cos y isi n y )  
 {
    cos v e
 sin y  
? 8
u
  e
 cos y ,     u
   e
 sin y
v
  e
 sin y ,   v
  e
 cos y
 
Hence, C-R equation satisfied. 
Qu e .  . a . S how | e
 | e
 . 
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Exercise  
Q pg  3 / S how that  f
  lim
    ,      f ( z h ) f ( z )
h
, f
  lim
    ,     f ( z i h ) f ( z )
h
 
Provided the limit exist. 
Sol
n
:-   Let  us suppose that limit exist. 
 Let z=x+iy, also h be real  
 ? z h  ( x h,y) , or ( x h) iy  
 Z+ih =(x, y+h)  or x+i (y+h) 
?   lim
    ,      f ( x h ,   y ) f ( x ,   y )
h
 lim
    ,      f ( z h ) f ( z )
h
     … … … . . ( i ) 
     lim
    ,      f ( x ,   y h ) f ( x ,   y )
h
  lim
    ,      f ( z ih ) f ( z )
h
  … … … . . ( ii ) 
Hence  f x &  f y  are given by eq
n
 (i)& (ii) when the limit exist. 
Que. 2  a. Show that  f(z)=x
2
+iy
2 
 is diff at all points on the line y=x 
              b. Show that it is nowhere analytic. 
Sol
n
:- given f(z)=x
2
+iy
2 
 
?   f(x, y ) x
2
+iy
2
 
 ?   f x=2x & f y = i2y =2iy 
    f x & f y   are cts everywhere  
The fun
c
  f  is diff if it satisfy C-R eq
n 
 
ie.    f y= if x  
      2iy=(2x)i 
 Ie.    y=x 
Hence, f(z) is diff at all pts on the line y=x. 
Qu e.   S upp ose  f  is an a ly ti c in a reg ion   f      there. S how that  f is const an t. 
Sol
n
:-   Let   f=u+iv be analytic in a region 
 ? Dif f  w . r . t. ‘x ’   ‘y ’ p art ia lly . 
 ? f x =u x+iv x 
  f y= u y+iv y 
 g iven f     
   {
u
  iv
      u
  i v
              u x=0, v x=0, u y=0, v y=0. 
  all the p artia l d eriv at ive are eq ual to z ero 
 Hence,  f  is constant. 
Que. 6  Assume that fun
c
 f is analytic in a region & that at the every point of the  
r egion, ei ther f     or  f    . S how that f is const an t. ,H in t consi d er f
2
.] 
Sol
n
 :-     the fu n
n
  f is analytic 
 ? f
2
 is also analytic 
C onsider  ( f
 ( z ) )
    f ( z ) f
 ( z )        ( if  f ( z )    or f  ( z )    ) 
       (f
2
)          z 
 ?  y Prev ious que , f
2
 is constant & hence  f . 
Que 8. Find all analytic function f = u + iv with u(x, y) = x
2
-y
2
. 
So l
  Giv en u ( x , y ) x
  y
  
? u
   x 
  u
    y 
we k now that C auch y Rie mann eq uati on is  
u
  v
  
? v
   x 
in te g r at e p artia lly w . r . t . y 
Free coaching of B.Sc (h) maths & JAM 
For more 8130648819 
 
? v  xy c 
? f ( x
  y
 ) i ( xy c ) 
f ( x iy )
  ic 
f ( z ) z
  c
         ,    c
   . 
Qu e  . S how that there is no an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
So l
  L et if p ossi b le there is an an aly ti c f uncti on 
f u iv  w it h u ( x , y ) x
  y
  
if the f uncti on f is an aly ti c then it sat i sf ie s C R eq uati on . 
?
 u
 x
  x  v
 y
 v  xy c
             … … . . ( ) 
  u
 y
  y   v
 x
 v   xy c
     … … . . (  ) 
Which is a contradiction. 
? we can not f in d an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
 lte r na te 
also , ( v
 )
        ( v
 )
              usin g ( i )   ( ii )  
b ut v
  
 v
  
    sin ce f is an aly ti c  
?      a contrad ict ion . 
en ti r e  an aly C R e q
  
Qu e   . S upp ose f is an en t ire f uncti on of the f or m 
  f ( x , y ) u ( x ) iv ( y ) 
S how that f is a line ar p oly nomia l . 
So l
  L et f ( x , y ) u ( x ) iv ( y ) is an en t ire f uncti on .        … ( i ) 
? it is an aly ti c ov er w hol e d isc . 
? it sat isfy C auch y Rie mann eq uati on 
 ote en ti r e an aly ti c C R eq uati on sa ti sf ie d 
So , u
  v
  
i . e .  u
 ( x ) v
 ( y )    , u
 ( x ) is only f uncti on of x  v
 ( y ) is the only f uncti on of y - 
 u
 ( x ) and v
 ( y ) b oth are consta nt . 
L et u
 ( x ) v
 ( y ) a  
now , {
  ( )    ( )   
in te g r at e 
u ( x ) ax c
  
v ( y ) ay c
  
Put the value of u & v in equation (i)  
?  f ( x , y ) ( ax c
 ) i ( ay c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 az b      w her e  b c
  i c
  
Hence, f  is a linear polynomial. 
Qu e .  . a . S how that e
  is en ti r e by v erif y in g the C auch y Rie mann eq uati on f or its r ea l and imagina r y p arts . 
or  Pro v e that e
  is an a ly ti c in the case of r ea l and imagina r y p art . 
b . Pro v e  
e
      e
  e
   
So l
  L et f ( z ) e
  e
    
 e
 . e
    
u iv e
 ( cos y isi n y )  
 {
    cos v e
 sin y  
? 8
u
  e
 cos y ,     u
   e
 sin y
v
  e
 sin y ,   v
  e
 cos y
 
Hence, C-R equation satisfied. 
Qu e .  . a . S how | e
 | e
 . 
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b . e
    f or any z  . 
c . e
  
 cis y 
d . e
    has in f in it ely many solu ti on f or any    . 
So l
   e
  is w ell d efie d f uncti on . 
e
  e
      e
 . e
  
 e
 ( cos  i sin ) 
? | e
 | v ( e
 )
  ( cos
 y sin
 y ) e
  
( b ) e
    f or any z   
as | e
 | e
    
 | e
 |    
 e
   . 
( c ) let x   
? e
  e
  
 cos y  i sin  cisy 
( d ) L et   r ( cos  isi n ) 
? e
  r e
   
e
    
 r e
   
 e
 e
  
 r e
   
 {
r e
  x log r                                                            and
 e
  
 e
   y  k      w her e k  ,  ,  , …
 
Since, k has infinitely many values. 
  has in f in it ely many v alues . 
 e
    has in f in it e ly many v alues p r ov id ed    . 
Qu e .  . F in d all the solu ti on of  
( a ) e
     (b) e
  i  (c) e
   3   (d) e
    i 
So l
  e
    
 e
    
   
 e
 e
  
   
 e
 ( cosy isi ny )   
 e
 cosy  , e
 sin y    …… . ( i)          on comparing real and imaginary part of both side 
squa r in g and ad d in g 
( e
 )
    
 e
    
 x log    
? f r om e q
 ( i ) 
cos     sin    
 y  k   k be any in te g er i . e . k   
? z x iy  ikz       w her e k   
i . e . z * ik  k be any in te g er + 
( b ) e
  i 
 e
    
 i 
 e
 . e
  
 i 
 e
 ( cosy isi ny ) i 
 e
 cosy  ,   e
 sin y         …. ( i) 
squa r in g and ad d in g 
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Exercise  
Q pg  3 / S how that  f
  lim
    ,      f ( z h ) f ( z )
h
, f
  lim
    ,     f ( z i h ) f ( z )
h
 
Provided the limit exist. 
Sol
n
:-   Let  us suppose that limit exist. 
 Let z=x+iy, also h be real  
 ? z h  ( x h,y) , or ( x h) iy  
 Z+ih =(x, y+h)  or x+i (y+h) 
?   lim
    ,      f ( x h ,   y ) f ( x ,   y )
h
 lim
    ,      f ( z h ) f ( z )
h
     … … … . . ( i ) 
     lim
    ,      f ( x ,   y h ) f ( x ,   y )
h
  lim
    ,      f ( z ih ) f ( z )
h
  … … … . . ( ii ) 
Hence  f x &  f y  are given by eq
n
 (i)& (ii) when the limit exist. 
Que. 2  a. Show that  f(z)=x
2
+iy
2 
 is diff at all points on the line y=x 
              b. Show that it is nowhere analytic. 
Sol
n
:- given f(z)=x
2
+iy
2 
 
?   f(x, y ) x
2
+iy
2
 
 ?   f x=2x & f y = i2y =2iy 
    f x & f y   are cts everywhere  
The fun
c
  f  is diff if it satisfy C-R eq
n 
 
ie.    f y= if x  
      2iy=(2x)i 
 Ie.    y=x 
Hence, f(z) is diff at all pts on the line y=x. 
Qu e.   S upp ose  f  is an a ly ti c in a reg ion   f      there. S how that  f is const an t. 
Sol
n
:-   Let   f=u+iv be analytic in a region 
 ? Dif f  w . r . t. ‘x ’   ‘y ’ p art ia lly . 
 ? f x =u x+iv x 
  f y= u y+iv y 
 g iven f     
   {
u
  iv
      u
  i v
              u x=0, v x=0, u y=0, v y=0. 
  all the p artia l d eriv at ive are eq ual to z ero 
 Hence,  f  is constant. 
Que. 6  Assume that fun
c
 f is analytic in a region & that at the every point of the  
r egion, ei ther f     or  f    . S how that f is const an t. ,H in t consi d er f
2
.] 
Sol
n
 :-     the fu n
n
  f is analytic 
 ? f
2
 is also analytic 
C onsider  ( f
 ( z ) )
    f ( z ) f
 ( z )        ( if  f ( z )    or f  ( z )    ) 
       (f
2
)          z 
 ?  y Prev ious que , f
2
 is constant & hence  f . 
Que 8. Find all analytic function f = u + iv with u(x, y) = x
2
-y
2
. 
So l
  Giv en u ( x , y ) x
  y
  
? u
   x 
  u
    y 
we k now that C auch y Rie mann eq uati on is  
u
  v
  
? v
   x 
in te g r at e p artia lly w . r . t . y 
Free coaching of B.Sc (h) maths & JAM 
For more 8130648819 
 
? v  xy c 
? f ( x
  y
 ) i ( xy c ) 
f ( x iy )
  ic 
f ( z ) z
  c
         ,    c
   . 
Qu e  . S how that there is no an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
So l
  L et if p ossi b le there is an an aly ti c f uncti on 
f u iv  w it h u ( x , y ) x
  y
  
if the f uncti on f is an aly ti c then it sat i sf ie s C R eq uati on . 
?
 u
 x
  x  v
 y
 v  xy c
             … … . . ( ) 
  u
 y
  y   v
 x
 v   xy c
     … … . . (  ) 
Which is a contradiction. 
? we can not f in d an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
 lte r na te 
also , ( v
 )
        ( v
 )
              usin g ( i )   ( ii )  
b ut v
  
 v
  
    sin ce f is an aly ti c  
?      a contrad ict ion . 
en ti r e  an aly C R e q
  
Qu e   . S upp ose f is an en t ire f uncti on of the f or m 
  f ( x , y ) u ( x ) iv ( y ) 
S how that f is a line ar p oly nomia l . 
So l
  L et f ( x , y ) u ( x ) iv ( y ) is an en t ire f uncti on .        … ( i ) 
? it is an aly ti c ov er w hol e d isc . 
? it sat isfy C auch y Rie mann eq uati on 
 ote en ti r e an aly ti c C R eq uati on sa ti sf ie d 
So , u
  v
  
i . e .  u
 ( x ) v
 ( y )    , u
 ( x ) is only f uncti on of x  v
 ( y ) is the only f uncti on of y - 
 u
 ( x ) and v
 ( y ) b oth are consta nt . 
L et u
 ( x ) v
 ( y ) a  
now , {
  ( )    ( )   
in te g r at e 
u ( x ) ax c
  
v ( y ) ay c
  
Put the value of u & v in equation (i)  
?  f ( x , y ) ( ax c
 ) i ( ay c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 az b      w her e  b c
  i c
  
Hence, f  is a linear polynomial. 
Qu e .  . a . S how that e
  is en ti r e by v erif y in g the C auch y Rie mann eq uati on f or its r ea l and imagina r y p arts . 
or  Pro v e that e
  is an a ly ti c in the case of r ea l and imagina r y p art . 
b . Pro v e  
e
      e
  e
   
So l
  L et f ( z ) e
  e
    
 e
 . e
    
u iv e
 ( cos y isi n y )  
 {
    cos v e
 sin y  
? 8
u
  e
 cos y ,     u
   e
 sin y
v
  e
 sin y ,   v
  e
 cos y
 
Hence, C-R equation satisfied. 
Qu e .  . a . S how | e
 | e
 . 
Free coaching of B.Sc (h) maths & JAM 
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b . e
    f or any z  . 
c . e
  
 cis y 
d . e
    has in f in it ely many solu ti on f or any    . 
So l
   e
  is w ell d efie d f uncti on . 
e
  e
      e
 . e
  
 e
 ( cos  i sin ) 
? | e
 | v ( e
 )
  ( cos
 y sin
 y ) e
  
( b ) e
    f or any z   
as | e
 | e
    
 | e
 |    
 e
   . 
( c ) let x   
? e
  e
  
 cos y  i sin  cisy 
( d ) L et   r ( cos  isi n ) 
? e
  r e
   
e
    
 r e
   
 e
 e
  
 r e
   
 {
r e
  x log r                                                            and
 e
  
 e
   y  k      w her e k  ,  ,  , …
 
Since, k has infinitely many values. 
  has in f in it ely many v alues . 
 e
    has in f in it e ly many v alues p r ov id ed    . 
Qu e .  . F in d all the solu ti on of  
( a ) e
     (b) e
  i  (c) e
   3   (d) e
    i 
So l
  e
    
 e
    
   
 e
 e
  
   
 e
 ( cosy isi ny )   
 e
 cosy  , e
 sin y    …… . ( i)          on comparing real and imaginary part of both side 
squa r in g and ad d in g 
( e
 )
    
 e
    
 x log    
? f r om e q
 ( i ) 
cos     sin    
 y  k   k be any in te g er i . e . k   
? z x iy  ikz       w her e k   
i . e . z * ik  k be any in te g er + 
( b ) e
  i 
 e
    
 i 
 e
 . e
  
 i 
 e
 ( cosy isi ny ) i 
 e
 cosy  ,   e
 sin y         …. ( i) 
squa r in g and ad d in g 
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( e
 )
    
 e
    
 x log    
? f r om eq uati on ( ) 
cos      , sin    
? y ( k  )
     w her e k be any in te g er 
? y 2 i ( k  )
   k be any in te g er 3 
( d ) e
    i 
 e
    
   i 
 e
 e
  
   i 
 e
 ( cos  i sin )   i 
 e
 cos       , e
 sin     …… ( i) 
squa r in g and ad d in g  
( e
 )
    
 e
  ( )
   
 x   log  
f r om eq uati on ( ) 
v cos   ,    v sin    
cosy  v     ,   sin y  v  
? y  k          k    be any in te g er 
? z {
  log  i . k    / k be any in te g er } 
( e ) e
   3 
 e
    
  3 
 e
 . e
  
  3 
 e
 ( cos  i sin )  3 
 e
 cos   3  ,   e
 sin      … . . ( i ) 
squa r in g and ad d in g . 
( e
 )
    
 e
  3             d oubt w hy d id we ta k e ve 3 , not 3 ? 
 x log 3 
From equation (1) 
3 cos   3 , 3 sin    
cosy       ,    sin y   
? y ( k  )     k be any in te g er 
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Exercise  
Q pg  3 / S how that  f
  lim
    ,      f ( z h ) f ( z )
h
, f
  lim
    ,     f ( z i h ) f ( z )
h
 
Provided the limit exist. 
Sol
n
:-   Let  us suppose that limit exist. 
 Let z=x+iy, also h be real  
 ? z h  ( x h,y) , or ( x h) iy  
 Z+ih =(x, y+h)  or x+i (y+h) 
?   lim
    ,      f ( x h ,   y ) f ( x ,   y )
h
 lim
    ,      f ( z h ) f ( z )
h
     … … … . . ( i ) 
     lim
    ,      f ( x ,   y h ) f ( x ,   y )
h
  lim
    ,      f ( z ih ) f ( z )
h
  … … … . . ( ii ) 
Hence  f x &  f y  are given by eq
n
 (i)& (ii) when the limit exist. 
Que. 2  a. Show that  f(z)=x
2
+iy
2 
 is diff at all points on the line y=x 
              b. Show that it is nowhere analytic. 
Sol
n
:- given f(z)=x
2
+iy
2 
 
?   f(x, y ) x
2
+iy
2
 
 ?   f x=2x & f y = i2y =2iy 
    f x & f y   are cts everywhere  
The fun
c
  f  is diff if it satisfy C-R eq
n 
 
ie.    f y= if x  
      2iy=(2x)i 
 Ie.    y=x 
Hence, f(z) is diff at all pts on the line y=x. 
Qu e.   S upp ose  f  is an a ly ti c in a reg ion   f      there. S how that  f is const an t. 
Sol
n
:-   Let   f=u+iv be analytic in a region 
 ? Dif f  w . r . t. ‘x ’   ‘y ’ p art ia lly . 
 ? f x =u x+iv x 
  f y= u y+iv y 
 g iven f     
   {
u
  iv
      u
  i v
              u x=0, v x=0, u y=0, v y=0. 
  all the p artia l d eriv at ive are eq ual to z ero 
 Hence,  f  is constant. 
Que. 6  Assume that fun
c
 f is analytic in a region & that at the every point of the  
r egion, ei ther f     or  f    . S how that f is const an t. ,H in t consi d er f
2
.] 
Sol
n
 :-     the fu n
n
  f is analytic 
 ? f
2
 is also analytic 
C onsider  ( f
 ( z ) )
    f ( z ) f
 ( z )        ( if  f ( z )    or f  ( z )    ) 
       (f
2
)          z 
 ?  y Prev ious que , f
2
 is constant & hence  f . 
Que 8. Find all analytic function f = u + iv with u(x, y) = x
2
-y
2
. 
So l
  Giv en u ( x , y ) x
  y
  
? u
   x 
  u
    y 
we k now that C auch y Rie mann eq uati on is  
u
  v
  
? v
   x 
in te g r at e p artia lly w . r . t . y 
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? v  xy c 
? f ( x
  y
 ) i ( xy c ) 
f ( x iy )
  ic 
f ( z ) z
  c
         ,    c
   . 
Qu e  . S how that there is no an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
So l
  L et if p ossi b le there is an an aly ti c f uncti on 
f u iv  w it h u ( x , y ) x
  y
  
if the f uncti on f is an aly ti c then it sat i sf ie s C R eq uati on . 
?
 u
 x
  x  v
 y
 v  xy c
             … … . . ( ) 
  u
 y
  y   v
 x
 v   xy c
     … … . . (  ) 
Which is a contradiction. 
? we can not f in d an aly ti c f uncti on f u iv w it h u ( x , y ) x
  y
 . 
 lte r na te 
also , ( v
 )
        ( v
 )
              usin g ( i )   ( ii )  
b ut v
  
 v
  
    sin ce f is an aly ti c  
?      a contrad ict ion . 
en ti r e  an aly C R e q
  
Qu e   . S upp ose f is an en t ire f uncti on of the f or m 
  f ( x , y ) u ( x ) iv ( y ) 
S how that f is a line ar p oly nomia l . 
So l
  L et f ( x , y ) u ( x ) iv ( y ) is an en t ire f uncti on .        … ( i ) 
? it is an aly ti c ov er w hol e d isc . 
? it sat isfy C auch y Rie mann eq uati on 
 ote en ti r e an aly ti c C R eq uati on sa ti sf ie d 
So , u
  v
  
i . e .  u
 ( x ) v
 ( y )    , u
 ( x ) is only f uncti on of x  v
 ( y ) is the only f uncti on of y - 
 u
 ( x ) and v
 ( y ) b oth are consta nt . 
L et u
 ( x ) v
 ( y ) a  
now , {
  ( )    ( )   
in te g r at e 
u ( x ) ax c
  
v ( y ) ay c
  
Put the value of u & v in equation (i)  
?  f ( x , y ) ( ax c
 ) i ( ay c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 a ( x iy ) ( c
  i c
 ) 
 az b      w her e  b c
  i c
  
Hence, f  is a linear polynomial. 
Qu e .  . a . S how that e
  is en ti r e by v erif y in g the C auch y Rie mann eq uati on f or its r ea l and imagina r y p arts . 
or  Pro v e that e
  is an a ly ti c in the case of r ea l and imagina r y p art . 
b . Pro v e  
e
      e
  e
   
So l
  L et f ( z ) e
  e
    
 e
 . e
    
u iv e
 ( cos y isi n y )  
 {
    cos v e
 sin y  
? 8
u
  e
 cos y ,     u
   e
 sin y
v
  e
 sin y ,   v
  e
 cos y
 
Hence, C-R equation satisfied. 
Qu e .  . a . S how | e
 | e
 . 
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b . e
    f or any z  . 
c . e
  
 cis y 
d . e
    has in f in it ely many solu ti on f or any    . 
So l
   e
  is w ell d efie d f uncti on . 
e
  e
      e
 . e
  
 e
 ( cos  i sin ) 
? | e
 | v ( e
 )
  ( cos
 y sin
 y ) e
  
( b ) e
    f or any z   
as | e
 | e
    
 | e
 |    
 e
   . 
( c ) let x   
? e
  e
  
 cos y  i sin  cisy 
( d ) L et   r ( cos  isi n ) 
? e
  r e
   
e
    
 r e
   
 e
 e
  
 r e
   
 {
r e
  x log r                                                            and
 e
  
 e
   y  k      w her e k  ,  ,  , …
 
Since, k has infinitely many values. 
  has in f in it ely many v alues . 
 e
    has in f in it e ly many v alues p r ov id ed    . 
Qu e .  . F in d all the solu ti on of  
( a ) e
     (b) e
  i  (c) e
   3   (d) e
    i 
So l
  e
    
 e
    
   
 e
 e
  
   
 e
 ( cosy isi ny )   
 e
 cosy  , e
 sin y    …… . ( i)          on comparing real and imaginary part of both side 
squa r in g and ad d in g 
( e
 )
    
 e
    
 x log    
? f r om e q
 ( i ) 
cos     sin    
 y  k   k be any in te g er i . e . k   
? z x iy  ikz       w her e k   
i . e . z * ik  k be any in te g er + 
( b ) e
  i 
 e
    
 i 
 e
 . e
  
 i 
 e
 ( cosy isi ny ) i 
 e
 cosy  ,   e
 sin y         …. ( i) 
squa r in g and ad d in g 
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( e
 )
    
 e
    
 x log    
? f r om eq uati on ( ) 
cos      , sin    
? y ( k  )
     w her e k be any in te g er 
? y 2 i ( k  )
   k be any in te g er 3 
( d ) e
    i 
 e
    
   i 
 e
 e
  
   i 
 e
 ( cos  i sin )   i 
 e
 cos       , e
 sin     …… ( i) 
squa r in g and ad d in g  
( e
 )
    
 e
  ( )
   
 x   log  
f r om eq uati on ( ) 
v cos   ,    v sin    
cosy  v     ,   sin y  v  
? y  k          k    be any in te g er 
? z {
  log  i . k    / k be any in te g er } 
( e ) e
   3 
 e
    
  3 
 e
 . e
  
  3 
 e
 ( cos  i sin )  3 
 e
 cos   3  ,   e
 sin      … . . ( i ) 
squa r in g and ad d in g . 
( e
 )
    
 e
  3             d oubt w hy d id we ta k e ve 3 , not 3 ? 
 x log 3 
From equation (1) 
3 cos   3 , 3 sin    
cosy       ,    sin y   
? y ( k  )     k be any in te g er 
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? z * log 3 i ( k  )  k be any in te g er + 
…… …… …… … …… …… … … … 
sin z  e
  
 e
   
 i
 
and cos z e
  
 e
   
  
cos ( ix ) e
 (  )
 e
  (  )
  e
   e
   
not e | sin x |     , | cosx |       x   
b ut sin ( z ) and cosz   are not b ound ed 
Complex sin & cos are not bdd 
in        sin x     cos x   
 cos ( ix ) e
   e
         ( as n  ) 
  | sin (  i ) |   | e
  
 e
   
|       
not e ( sin z )
  d
dz
[
  i
( e
  
 e
   
) ] 
   i
[ i e
  
 ( i ) e
   
] 
   [ e
  
 e
   
] 
 cos z  
Qu e   . V erif y the id en ti ti es  
( a ) sin z  sin z cos z   ( b ) sin
 z cos
 z     ( c ) ( sin z )
  cosz 
( a ) we k now that sin z  e
  
 e
   
 i
 
and cos z e
  
 e
   
  
? {
LHS sin z   i
( e
    e
    )
RHS  sin cos                           
  6 4
e
  
 e
   
 i
5 4
e
  
 e
   
 5 7   i
( e
    e
    ) 
? sin z  sin z cosz  
 Qu e   . F in d ( cos )
  
So l
  d
dz
( cos ) d
dz
4
  ( e
  
 e
   
) 5   ( i e
  
 ie
   
) i
 ( e
  
 e
   
)  
Multiplying & dividing by i, we get 
 i
  i
( e
  
 e
   
)    i
( e
  
 e
   
)  sin  
Qu e   . Pro v e that sin ( x iy ) sin cos hy i cos sin hy 
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27 docs|150 tests

FAQs on Analytic Function (Solved Questions) - Topic-wise Tests & Solved Examples for Mathematics

1. What is an analytic function?
An analytic function is a complex-valued function that is locally given by a convergent power series. It is a function that can be expressed as a power series that converges in a neighborhood of each point in its domain.
2. What are some properties of analytic functions?
Some properties of analytic functions include the fact that they are infinitely differentiable, they preserve angles between curves, and their real and imaginary parts satisfy the Cauchy-Riemann equations. Additionally, the derivative of an analytic function is also analytic.
3. How can I determine if a function is analytic?
To determine if a function is analytic, one can check if it satisfies the Cauchy-Riemann equations. These equations relate the partial derivatives of the function's real and imaginary parts. If the partial derivatives satisfy these equations, then the function is analytic.
4. Can an analytic function have singularities?
Yes, an analytic function can have singularities. A singularity occurs when a function is not analytic at a certain point. There are different types of singularities, such as removable singularities, poles, and essential singularities. These occur when the function behaves in a certain way at the singularity point.
5. What are some applications of analytic functions?
Analytic functions have various applications in mathematics and physics. They are used in complex analysis to study properties of functions and their behavior. In physics, they are used to model and solve problems in fluid dynamics, electromagnetism, and quantum mechanics. Analytic functions also find applications in engineering, signal processing, and image processing.
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