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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
Page 2
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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
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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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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
Page 4
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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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( 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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