CONSTRUCTION OF ANALYTIC FUNCTION:
MILNE THOMSON METHOD:
Let f(z)=u+iv is to be constructed
 If R.P u is given ∂u/∂x, ∂u/∂y can be calculated ,
f(z)=∫( ∂u/∂x(z,0) - ∂u/∂y(z,0)) dz
 If I.P v is given,
f(z)=∫ ∂v/∂y(z,0) +i∂v/∂x(z,0)) dz
PROBLEMS:
1. Given that u=sinx / (cosh2x-cos2x) find the analytic function whose real
part is u.
Solution:
f(z)=∫ɸ1(z,0) dz – i ∫ɸ2(z,0) dz
∂u/∂x = [(cosh2y-cos2x) ( 2cos2x) –(sin2x) (2sin2x)]/[cosh2y-cos2x]2
= [ 2cos2x.cosh2y-2cos22x-2sin22x] / [cosh2y-cos2x]2
= [2cos2x.cosh2y-2] / [cosh2y-cos2x]2
∂u/∂x(z,0)=2cos2z-2 / (1-cos2z)2
=-2/1-cos2z
=-1/[(1-cos2z)/2]
= -1/sin2z
=-cosec2z
∂u/∂y = [(cosh2y-cos2x) (0) –sin2x2sinh2y] / [cosh2y-cos2x]2
= (-sin2x sinh2y) / (cosh2y-cos2x)2
∂u/∂y(z,0)=-2sin2x(0) / (cosh2y-cos2x)2 =0
f(z)=∫(-cosec2z – i(0))dz
=-∫cosec2z dz
=cot z +c
CONSTRUCTION FOR ANALYTIC FUNCTION:
MILNE-THOMSON METHOD:
Case 1:
To find f(z), when u is given,
f’(z)=ux+ivx
Case 2:
= ux-ivy
f’(z)= ux(x,y) –ivy(x,y)
∫ f’(z)dz=∫ux(z,0) dz – i ∫uy(z,0) dz +c
f(z)=∫ux(z,0)dx – i ∫uy(z,0) dz +c
To find f(z), when v is given,
f’(z)=ux+ivx
= vy+ivx
∫ f’(z)dz=∫vy(z,0)dz + i ∫vx(z,0) dz +c
f(z)=∫vy(z,0)dz + i ∫ vx(z,0) dz +c
PROBLEMS:
1. Determine the analytic function whose real part is x 3-3xy2+3x2-3y2+1
Solution:
u=x3-3xy2+3x2-3y2+1
f(z)=∫ux(z,0) dz – i ∫uy(z,0) dz +c
ux(x,y)=3x2-3y2+6x,
uy=-6xy-6y
2
ux(z,0)=3z +6z,
uy(z,0)=0
2
f(z)=∫(3z +6z)dz+c
=z3+3z2+c
2. Determine the analytic function where u=e2x
Solution:
f(z)=∫ux(z,0)dz – i ∫uy(z,0)dy +c
ux(x,y)=2e2x.sin2y
ux(z,0)=2e2z.sin 0
=0
uy(x,y)=e2xcos2y.2
=2e2xcos2y
uy(z,0)=2e2zcos0
=2e2z
f(z)=0-i∫2e2z/2 +c
=ie2z+c
3. Determine the analytic function f(z) such that u-v=ex(cosy-siny)
Solution:
f(z)=u+iv
i f(z)=i (u+iv)
=iu-v
f(z) + i f(z)= u+iv+iu-v
f(z) (1+i)=(u-v) + i(u+v)
f(z)=u+iv
Given u-v = u = ex(cosy-siny)
ux=ex(cosy-siny)
ux(z,0)=ez(cos0-sin0)
=ez
uy=ex(-siny-cosy)
uy(z,0)=-ex
f(x,y)=∫ux(z,0)dz – i ∫uy(z,0)dz +c
f(x,y)=∫ezdz – i ∫-ezdz +c
f(z)(1+i) =ez+i ez +c
f(z)=[ez(1+i)] / [1+i] + [c] / [1+i]
f(z)=ez+c
4. Determine the analytic function f(z) such that u+v=x/x 2+y2 subject to the
condition f(1)=1
Solution:
f(z)=u+iv
i f(z)=i(u+iv)
=iu-v
f(z)(1+i)=u-v +i(u+v)
f(z)=u+iv, u=u-v, v=u+v
given,
u+v=v=x/x2+y2
vx=(x2+y2) (1) –x(2x) / (x2+y2)2
vx(x,y)=y2-x2 / (x2+y2)
vx(z,0)=-z2 / (z2+0)2
= -z2 / z4
=-1/z2
vy=(x2+y2)(0) –x(2y) / (x2+y2)2
= -2xy / (x2+y2)2
vy(z,0)=0
f(z)=∫ vy(z,0)dz + i ∫vx(z,0)dz +c
=i ∫-1/z2dz +c
=-i(z-1/-1) +c
=i/z +c
5. Determine the analytic function f(z) such that 2u+3v=sin2x / cosh2y-cos2x
Solution:
f(z)=u+iv
2f(z)=2u+i2v
i f(z)=iu-v
3i f(z)=i3u-3v
2f(z)-3if(z)=2u+i2v-i3u+3v
=2u+3v+i(2v-3u)
f(z)(2-3i)=u+iv
f(z)=u+iv
u=2u+3v,
v=2v-3u
2u+3v=u=sin2x / cosh2y-cos2x
ux=(cosh2y-cos2x)2cos2x – sin2x(2sin2x) / (cosh2y-cos2x)2
ux(z,0)=(cosh2(0)-cos2z) 2cos2x – sin 2x(2sin2x) / (cosh2y-cos2x)2
=-cosec2z
uy=0-sin2x(2sinh2y) / (cosh2y-cos2x)2
uy(z,0)=0
f(z)=∫ ux(z,0)dz – i ∫uy(z,0)dz +c
= - ∫cosec2zdz+c
f(z)(2-3i)=cotz +c
f(z)=cotz/ (2-3i) + c/(2-3i)
=cotz(2+3i) / (2-3i)(2+3i) +c
= cotz (2+3i) / (2-3i)(2+3i) +c
=cotz(2+3i)/13 +c
BILINEAR TRANSFORMATION OR MOBIUS TRANSFORMATION:
W=az+b / cz+d , ad-bc≠0
Where a,b,c,d are complex numbers is called bilinear transformation.
NOTE:
The bilinear transformation which transforms z1,z2,z3 into w1,w2,w3 is…,
[[(w-w1)(w2-w3)] / [(w-w3)(w2-w1)]] = [[ (z-z1) (z2-z3)] / [(z-z3) (z2-z1)]]
PROBLEMS:
1. Find the bilinear transformation that mps the points z=-i,0,i, into w=-1,i,1
Solution:
z1=-i,z2=0,z3=i
w1=-1, w2=i,w3=1
[[(w-w1)(w2-w3)] / [(w-w3)(w2-w1)]] = [[ (z-z1) (z2-z3)] / [(z-z3) (z2-z1)]]
[[(w+1)) (i-1)] / [(w-1)(i+1))]] = [[ (z-i) (0-i)] / [(z-i) (0+i)]]
[[(w+1)) (i-1)] / [(w-1)(i+1))]] = -[[(z+i)(i)] / [(z-i)(i)] ]
[[(w+1) (i-1)(i-1)] / [(w-1) (i+1) (i-1)] ] = -(z+i) / (z-i)
[(w+1)(i2-2i+1)] /[(w-1) (1+1) ] = -(z+i) / (z-i)
(w+1)(-2i) / (w-1)2 = -(z+i) / (z-i)
(w+1) (-i) / (w-1) = -(z+i) / (z-i)
(w+1) / (w-1) = (z+i) (-i) / (z-i) (i)
=(z+i)(-i) / (z-i)(-i2)
= (-zi+1) / (z-i)
=-zi+i / z-i
w+1+w-1 / w+1-w+1 = -zi+1+z-i / -zi+1-z+i
2w/2= -i(z+1) + (1+z) / -z(i+1) + (1+i)
w=(z+1)(1-i) / (1+i)(1-z)
= (z+1)(1-i) (1-i) / (1-z) (1+i) (1-i)
= (z+1) (1-2i+1) / (1-z) (1+1)
= (z+1) (-i) / (1-z)
=-i(z+1) / (1-z)
2. Find the bilinear transformation maps the points 0,1,∞, into -5,-1,3
respectively
Solution:
By bilinear transformation:
[[(w-w1)(w2-w3)] / [(w-w3)(w2-w1)]] = [[ (z-z1) (z2-z3)] / [(z-z3) (z2-z1)]]
z1=0,z2=1,z3=∞
w1=-5, w2=-1,w3=3
(w+5) (-1-3) / (w-3) (-1+5) = (z-0) (1-∞) / (z-∞)(1-0)
(w+5) (-4) / (w-3) (4) = z(1-∞) / (z-∞)
= z(-∞) (1/∞ − 1) / (-∞) (z/∞-1)
=z(-1) / (-1)
=z
-(w+5) / (w-3)= z
-(w+5)=z(w-3)
-w-5=zw-3z
w-zw=5-3z
-w(1+z)=5-3z
To find invariant point
w=(3z-5) / (1+z)
w=z
z=(3z-5) / (1+z)
(z+1)z=3z-5
z2+z=3z-5
z2-2z+5=0
z=2±√(4-4(5) / 2
= 2±4i / 2
=1±2i