UUnit4-VM
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Transcript of UUnit4-VM
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UNIT-IV
Transformations in the complex plane:
Consider the complex valued function
W=f(z)---------(1)
A complex number z=x+iy determines a point P(x,y) in the complex plane
and is referred as the point z. The point w= u+iv, is represented by a point
Q(u,v) in theu-v plane. Thus w= f(z) represents a transformation and
transforms a point P(x,y) to a unique point Q(u,v) in the complex plane.
Conformal Transformation:
The transformation w= f(z) transforms the curves C1 and C2 to the curves
C11 and C2
1 and intersects at a given point then the transformation is said
to be a conformal transformation.
I):The transformation w=1/z
f(z)= 1/z is analytic with
As such, this transformation is conformal at every point . The transformation
is carried by taking in polar form. Then from
the above Equation so that a point is transformed to the point
Ex: show that the transformation
w = 1/z transforms a circle to a circle or a straight line. Using W = u+iv, z= x+iy. And w = f(z) gives
equating real and imaginary parts gives.
----------(I)
Let us consider any circle in z-plane . Its Cartesian equation is of the form
0.zfor z
-1z)f
2
1 = ,(
iiRewandrez ==
( ) ( ) r,-R, /1=( )r,
22 vu
iv-u
ivu
1iyx
+=
+=+
,22
vu
ux
+= ,
22vu
-vy
+=
0c2fy2gxyx 22 =++++
-
Substituting for x and y from equation (I), we get
-------(II)
The above equation represents a circle in the w- plane if ,
and a straight line if c =0.
(II) :Transformation w =
1.Consider the transformation w= --(I)
The transformation is conformal for .
Now u+iv =
So, that
Let . .Which represents a rectangular
Hyperbola. V= constant. V=2B, Which is also a rectangular hyperbola.
The two families of curves
Under the given transformation w = the rectangular hyperbolas
in the z-plane transforms to the st-lines u=A, and v=2B, in the w-plane.
2.Now consider a line parallel to y-axis. The equation Of this is of the form
x=a, where a- is a constant.
Then
The equation represents equation of parabola in the w-plane having vertex at
( ) 012fu- 2guvuc 22 =+++0c
2z
2z0z
( ) i(2xy)yxiyx 22 +=+ 2
2xyvyxu 22 == ,
Au i.eAyx 22 ==
lyorthogonalintersect Bxy and Ayx 22 ==
2z Bxy and Ayx
22 ==
uyaor uyx 2222 ==
v 2ayor v2xy ==
)1()(4 = 222 a-uav
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the point , and its axis is the negative u-axis.
3.Again consider a line parallel to the x-axis. Its equation is of the form
y=b, where b is constant.
V=2xy or v=2xb
Which represents a parabola in the w-plane having vertex at the point
and its axis is positive u-axis.
Hence the transformations W= transforms st-lines parallel to y-axis
to parabolas having the negative u-axis as their common axis and the straight
lines parallel to x- axis to parabolas having the positive u-axis as their
common axis.
(III) :Transformation:
Here for any z. Therefore the transformation is
conformal for all z.
-----(1)
We shall find the image in the w-plane corresponding to the straight
lines parallel to the co-ordinate axes in the z-plane. Let x = constant
y= constant.
Squaring and adding equations (1) we get
)0,2(a
2bor == 222 xu yxu
)(4) 2222 bubbb 4(u +=+=
( )0,2b-2z
zew =
0( = z1 ez)f
isiny)cosyee ivu xiyx +==+ + (
sinyevcosyeu xx == 2x22 evu =+
-
and by dividing
Case-1:Let x = c Where c is a constant.
This represents a circle with center origin and radius r, in the w-plane.
Case-2: Let y= c where c- is a constant
This represents a st-line passing through origin in the w-plane.
Conclusion : The st-line parallel to the x-axis in the z-plane maps onto
a st-line passing through the origin in the w-plane. The st-line parallel to
y-axis in the z-plane maps onto a circle with center origin and radius r.
A tangent is drawn at the point of intersection of these two curves in the w-
plane, the angle subtended is 90. Hence the two curves are orthogonal
trajectories of each other.
(IV):The Transformation :
Here The transformation is conformal at all points except
at 0 and . The transformation is also known as the Joukowskis
transformation.
Equating real and imaginary parts
After simplifying
2x22 evu =+
tanyu
v=
( ) 2c2c22 reevu ===+ 2
muvor mtancu
v===
z
azw
2
+=
( )z
azf
21 = 1
a
irezLet = i-2
i er
areivu +=+
( ) isin-cosr
a)isinr(cos
2
++=
-
-------(1)
Squaring and adding
Case-1: When r = constant then above eqn becomes
Which represents an ellipse in w-plane with foci
Hence the circle in the z-plane maps onto an
ellipse in the w-plane with foci
2. Eliminating r in the equation (1).
or
Now represents a circle with centre origin and radius r
in the z-plane.
This represents a st-line in the z-plane passing through origin.
Now the equation (2) becomes
Where A=2 acos , B=2a sin
( ) ( )ra-rv
sinrar
u22 /
:/
cos =+
=
( ) ( )1
//22
=+
+ ra-r
v
rar
u
2
2
2
2
1=+2
2
2
2
b
v
a
u
)0,( 22 ba
constant,rz ==
a,0)2(
2
2
2
2
2
asin
v
cos
u4=
1
)) 22=
(2asin
v
(2acos
u22
ire=z
tanx
y
x
ytan-1 =
= and
1=2
2
2
2
B
v
A
u
-
This represents a hyperbola in the w-plane with foci.
The both conics (ellipse & hyperbola) have the same foci, independent of
r and and they are called confocal conics .
(V):Bilinear Transformation:
Let a,b,c, and d be complex constants such that
ad bc 0. Then the transformation defined by,
is called bilinear transformation. Solving for z, we find
Which is called the inverse bilinear transformation.
The transformation (1) establishes one-one correspondence between the
points in the z-and w- plane.
Now from equation (1)
Since the above equation is a quadratic equation there exists exactly two
such points for a given transformation. These are called the fixed points
or invariant points of the transformation.
Note 1:
There exists a bilinear transformation that maps three given distinct
points Onto three given distinct points
a,0)BA( 22 2()0,
)1(+
+=
dcz
bazw
)2(
+=
acw
b-dwz
0b-a)z(dczdcz
bazw 2 =+
+
+= or
321 zzz ,,
yrespectivl www 32,1,
-
Solving this equation for w in terms of z,
we obtain the bilinear transformation that
transforms
Ex: Find bilinear transformation that maps the points 1,i,-1 on to the
points i,0,-1 respectively.
Under this transformation find the image of
Also find the invariant points of this transformation.
Using the formula
We get
To find the image of
we rewrite the above equations as
If and using this condition we get u>0
Under this transformation the image of is u>0
Which is right half of w plane
( )( )( )( )
=
123
321
www-w
wwww ( )( )( )( )123
321
zzw-z
zzzz
to onzzz 321 ,, yrespectivl www 32,1,
1z