Conjugate Harmonic Function: (or) Harmonic Conjugate Function:
If f(z) = u+iv be an analytic functions and u, v are harmonic functions, then u and v are said
to be conjugate harmonic functions.
u v
Note: We have f 1 ( z) i
x x
u u
By C-R equations f 1 ( z ) i
x y
v v
f 1 ( z) i
y x
v u
f 1 ( z) i
y y
Finding harmonic conjugate of an analytic function when other part is given:
Milne – Thomson’s Method
Case I: When real part u = u(x,y) given, to find analytic function f(z) and imaginary
part v = v(x,y):
u u
Procedure: Step1: Consider f 1 ( z ) i
x y
Step2: Put x = z and y = 0 we get f 1 ( z ) in z.
Step3: Integrate f 1 ( z ) w.r.t. z. We get analytic function f(z).
Step4: To get imaginary part ‘v’ Put z = x+iy in f(z) and express as u+iv form.
Case II: When imaginary part v = v(x,y) given, to find analytic function f(z) and real
part u = u(x,y):
v v
Procedure: Step1: Consider f 1 ( z ) i
y x
Step2: Put x = z and y = 0 we get f 1 ( z ) in z.
Step3: Integrate f 1 ( z ) w.r.t. z. We get analytic function f(z).
Step4: To get real part ‘u’ Put z = x+iy in f(z) and express as u+iv form.
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Dr. K.S. Balamurugan, RVR & JC College of Engineering, Guntur, A.P.
Problems:
1. Determine the conjugate harmonic and analytic function whose real part is
x 3 3xy 2 3x 2 3 y 2 1
Sol: Given u x 3 3xy 2 3x 2 3 y 2 1
u u
Consider f 1 ( z ) i
x y
f 1 ( z ) (3x 2 3 y 2 6 x) i (6 xy 6 y )
Put x = z and y = 0
f 1 ( z ) 3z 2 6 z
Integrating f ( z ) z 3 3z 2 c which is an analytic function.
Put z = x+iy
f ( z ) ( x iy) 3 3( x iy) 2 c
f ( z ) ( x 3 iy 3 3x 2 iy 3xy 2 3( x 2 y 2 i 2 xy) c
f ( z ) ( x 3 3xy 2 3x 2 3 y 2 ) i ( y 3 3x 2 y 6 xy) c
Let c 1 ic1
f ( z ) ( x 3 3xy 2 3x 2 3 y 2 1) i ( y 3 3x 2 y 6 xy c1 )
Imaginary part v y 3 3x 2 y 6 xy c1
1
2. Show that u log( x 2 y 2 ) is harmonic and find its harmonic conjugate function.
2
1
Sol: Given u log( x 2 y 2 )
2
u 1 1 x 2 u ( x 2 y 2 ) x(2 x) y 2 x 2
x 2 x y
2x 2 and x 2 ( x 2 y 2 ) 2 2 2 2
x y2 (x y )
2 2
u 1 1 y 2 u ( x 2 y 2 ) y (2 y ) x 2 y 2
2y 2 2 2 2
y 2 x 2 y 2 x y 2 and y 2 (x2 y 2 )2 (x y )
2u 2u y2 x2 x2 y2
0
x 2 y 2 ( x 2 y 2 ) 2 ( x 2 y 2 ) 2
u satisfies Laplace equation, hence u is harmonic function.
u u
Consider f 1 ( z ) i
x y
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Dr. K.S. Balamurugan, RVR & JC College of Engineering, Guntur, A.P.
x y
f 1 ( z) i 2
x y
2 2
x y2
Put x = z and y = 0
1
f 1 ( z)
z
Integrating f ( z ) log z c which is an analytic function.
Put z = x+iy
f ( z ) log(x iy) c
log(rei ) c
log r log e i c
log x 2 y 2 i c
log x 2 y 2 i tan 1 ( y / x) c
log x 2 y 2 i tan 1 ( y / x) c
1
log( x 2 y 2 ) i tan 1 ( y / x) c
2
Imaginary part v tan 1 ( y / x) c
3. Determine the conjugate harmonic and analytic function whose imaginary part is
sinh x sin y
Sol: Given v sinh x sin y
v v
Consider f 1 ( z ) i f 1 ( z ) sinh x cos y i cosh x sin y
y x
Put x = z and y = 0 f 1 ( z ) sinh z
Integrating f ( z ) cosh z c which is an analytic function.
Put z = x+iy
f ( z ) cosh(x iy) c
We have cos(i ) cosh , sin(i ) i sinh
f ( z ) cos i ( x iy) c
cos(ix y ) c
(cosix) cos y (sin ix) sin y c cosh x cos y i sinh x sin y c
Hence real part u cosh x cos y c .
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Dr. K.S. Balamurugan, RVR & JC College of Engineering, Guntur, A.P.
4. If u = ex [(x2- y2) cos y – 2xy sin y] is a real part of an analytic function find the analytic
function.
Sol: Given u = ex [(x2-y2) cos y – 2xy sin y]
= ex [ (x2-y2) cos y – 2xy sin y] + ex [ 2x cos y – 2y sin y]
= ex [-2y cos y - (x2-y2)sin y - 2x sin y -2xy cos y]
u u
Consider f 1 ( z ) i
x y
= ex [ (x2-y2) cos y – 2 x y sin y + 2x cos y – 2y sin y ]
– iex [ -2y cos y –(x2-y2) siny – 2x sin y – 2xy cos y]
Put x = z and y = 0
We get f 1 ( z ) z2 ez + 2zez
= ez (z2 + 2z)
e f ( x) f
( x) dx e x f ( x)
x 1
Integrating on both sides we have
f(z) = ez z2 + c (or) f(z) = ez z2 + ic
sin 2 x
5. Find the analytic function whose real part is
cosh 2 y cos 2 x
sin 2 x
Sol: Given real part u
cosh 2 y cos 2 x
u u
Consider f 1 ( z ) i
x y
Put x = z and y = 0
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Dr. K.S. Balamurugan, RVR & JC College of Engineering, Guntur, A.P.
6. Find the analytic function u–v = (x – y) (x2 + y2+ 4xy).
Sol:
Note: f(z) = u + iv - (1)
i f (z) = iu – v - (2)
(1)+(2) (1+i) f (z) = (u-v) + i (u+v)
Let F(z) = U+ i V
Where F(z) = (1+i) f (z), U = (u-v) and V = (u+v)
Given real part U = (x – y) (x2 + y2+ 4xy)
U U
Consider F 1 ( z ) i
x y
= [(x – y) (2x+ 4y)+ (x2 + y2+ 4xy)]-i [(x – y) (2y+4x)- (x2 + y2+ 4xy)]
Put x = z and y = 0
F 1 ( z ) [(z) (2z)+ (z2 )]-i [(z) (4z)- z2 )] = 3z2-i 3z2 = 3z2(1-i )
Integrating F ( z ) z 3 (1 i) c
put F(z) = (1+i) f (z),
(1 i ) f ( z ) z 3 (1 i ) c
1 i c (1 i ) 2 2i
f ( z) z 3 z3 c1 z 3 c1 iz 3 c1
1 i 1 i (1 i )(1 i ) 2
c
Which is required analytic function f ( z ) iz 3 c1 where c1
1 i
Home Assignment:
(9) Determine the analytic function and imaginary part if real part is Cosx Coshy.
(10) Find the regular function whose imaginary part is e x sin y.
cos sin x e y
(11) Determine the analytic function f(z) = u+iv if u v and f ( / 2) 0
2(cos x cosh y )
2 sin 2 x
(12) Determine the analytic function f(z) = u+iv if u v .
e 2y
e 2 y 2 cos 2 x
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Dr. K.S. Balamurugan, RVR & JC College of Engineering, Guntur, A.P.
