1. The document discusses various types of transformations in complex analysis, including translation, rotation, stretching, and inversion.
2. Under inversion (1/w=z), a straight line is mapped to a circle if it does not pass through the origin, and to another straight line if it does pass through the origin. A circle is always mapped to another circle.
3. A general bilinear or Möbius transformation can be expressed as a combination of translation, rotation, stretching, and inversion.
Integration Made Easy!
The derivative of a function can be geometrically interpreted as the slope of the curve of the mathematical function f(x) plotted as a function of x. But its implications for the modeling of nature go far deeper than this simple geometric application might imply. After all, you can see yourself drawing finite triangles to discover slope, so why is the derivative so important? Its importance lies in the fact that many physical entities such as velocity, acceleration, force and so on are defined as instantaneous rates of change of some other quantity. The derivative can give you a precise intantaneous value for that rate of change and lead to precise modeling of the desired quantity.
Integration Made Easy!
The derivative of a function can be geometrically interpreted as the slope of the curve of the mathematical function f(x) plotted as a function of x. But its implications for the modeling of nature go far deeper than this simple geometric application might imply. After all, you can see yourself drawing finite triangles to discover slope, so why is the derivative so important? Its importance lies in the fact that many physical entities such as velocity, acceleration, force and so on are defined as instantaneous rates of change of some other quantity. The derivative can give you a precise intantaneous value for that rate of change and lead to precise modeling of the desired quantity.
this is the ppt on application of integrals, which includes-area between the two curves , volume by slicing , disk method , washer method, and volume by cylindrical shells,.
this is made by dhrumil patel and harshid panchal.
These slides contain information about Euler method,Improved Euler and Runge-kutta's method.How these methods are helpful and applied to our questions are detailed discussed in the slides.
this is the ppt on application of integrals, which includes-area between the two curves , volume by slicing , disk method , washer method, and volume by cylindrical shells,.
this is made by dhrumil patel and harshid panchal.
These slides contain information about Euler method,Improved Euler and Runge-kutta's method.How these methods are helpful and applied to our questions are detailed discussed in the slides.
An ellipse is the locus of a point which moves in such a way that its distance form a fixed point is in constant ratio to its distance from a fixed line. The fixed point is called the focus and fixed line is called the directrix and the constant ratio is called the eccentricity of a ellipse denoted by (e).
In other word, we can say an ellipse is the locus of a point which moves in a plane so that the sum of it distances from fixed points is constant.
2.1 Standard Form of the equation of ellipse
Let the distance between two fixed points S and S' be 2ae and let C be the mid point of SS.
Taking CS as x- axis, C as origin.
Let P(h,k) be the moving point Let SP+ SP = 2a (fixed distance) then
(ii) Major & Minor axis : The straight line AA is called major axis and BB is called minor axis. The major and minor axis taken together are called the principal axes and its length will be given by
Length of major axis 2a Length of minor axis 2b
(iii) Centre : The point which bisect each chord of an ellipse is called centre (0,0) denoted by 'C'.
(iv) Directrix : ZM and Z M are two directrix and their equation are x= a/e and x = – a/e.
(v) Focus : S (ae, 0) and S (–ae,0) are two foci of an ellipse.
(vi) Latus Rectum : Such chord which passes through either focus and perpendicular to the major axis is called its latus rectum.
Length of Latus Rectum :
If L is (ae, 𝑙 ) then 2𝑙 is the length of
SP+S'P=
{(h ae)2 k 2} +
= 2a
Latus Rectum.
Length of Latus rectum is given by
2b2
.
h2(1– e2) + k2 = a2(1– e2)
Hence Locus of P(h, k) is given by. x2(1– e2) + y2 = a2(1– e2)
2
a
(vii) Relation between constant a, b, and e
a 2 b2
b2 = a2(1– e2) e2 =
a 2
x2
a 2 +
y
a 2 (1 e 2 ) = 1
e =
a 2
Result :
Major Axis
(a) Centre C is the point of intersection of the axes of an ellipse. Also C is the mid point of AA.
(b) Another form of standard equation of ellipse
x 2 y2
a 2 b2
1 when a < b.
Directrix Minor Axis Directrix x = -a/e x = a/e
Let us assume that a2(1– e2 )= b2
The standard equation will be given by
x2 y2
a2 b2
2.1.1 Various parameter related with standard ellipse :
In this case major axis is BB= 2b which is along y- axis and minor axis is AA= 2a along x- axis. Focus S(0,be) and S(0,–be) and directrix are y = b/e and y = –b/e.
2.2 General equation of the ellipse
The general equation of an ellipse whose focus is (h,k) and the directrix is the line ax + by + c = 0 and the eccentricity will be e. Then let P(x1,y1) be any point on the ellipse which moves such that SP = ePM
Let the equation of the ellipse x
y2
a > b
(x –h)2 + (y –k)2 =
e 2 (ax1 by1 c) 2
a 2 b2
1 1 a 2 b2
(i) Vertices of an ellipse : The point of which ellipse cut the axis x-axis at (a,0) & (– a, 0) and y- axis at (0, b) & (0, – b) is called the vertices of an ellipse.
Hence the locus of (x1,y1) will be given by (a2 + b
2024.06.01 Introducing a competency framework for languag learning materials ...Sandy Millin
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
How to Make a Field invisible in Odoo 17Celine George
It is possible to hide or invisible some fields in odoo. Commonly using “invisible” attribute in the field definition to invisible the fields. This slide will show how to make a field invisible in odoo 17.
The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
For more information, visit-www.vavaclasses.com
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
This is a presentation by Dada Robert in a Your Skill Boost masterclass organised by the Excellence Foundation for South Sudan (EFSS) on Saturday, the 25th and Sunday, the 26th of May 2024.
He discussed the concept of quality improvement, emphasizing its applicability to various aspects of life, including personal, project, and program improvements. He defined quality as doing the right thing at the right time in the right way to achieve the best possible results and discussed the concept of the "gap" between what we know and what we do, and how this gap represents the areas we need to improve. He explained the scientific approach to quality improvement, which involves systematic performance analysis, testing and learning, and implementing change ideas. He also highlighted the importance of client focus and a team approach to quality improvement.
Students, digital devices and success - Andreas Schleicher - 27 May 2024..pptxEduSkills OECD
Andreas Schleicher presents at the OECD webinar ‘Digital devices in schools: detrimental distraction or secret to success?’ on 27 May 2024. The presentation was based on findings from PISA 2022 results and the webinar helped launch the PISA in Focus ‘Managing screen time: How to protect and equip students against distraction’ https://www.oecd-ilibrary.org/education/managing-screen-time_7c225af4-en and the OECD Education Policy Perspective ‘Students, digital devices and success’ can be found here - https://oe.cd/il/5yV
How to Split Bills in the Odoo 17 POS ModuleCeline George
Bills have a main role in point of sale procedure. It will help to track sales, handling payments and giving receipts to customers. Bill splitting also has an important role in POS. For example, If some friends come together for dinner and if they want to divide the bill then it is possible by POS bill splitting. This slide will show how to split bills in odoo 17 POS.
2. 2
CONFORMAL MAPPING
Content :
• Transformation
• Type of transformation
• Bilinear or Mobius transformation
• Circle under bilinear transformation
• Cross ratio
Transformation :
We know that w f(z)= for complex plane
u iv f (x iy)⇒ + = +
u u(xy)and v v(xy)⇒ = =
These equation define transformation or mapping from z-plane into w-plane the points z and w are called
the image of each after as shown in fig.
Here we write i
z re θ
= and i
w e φ
= ρ
CONFORMAL MAPPING
A mapping that preserves angle between any oriented curves both in magnitude and sense is a
conformal mapping.
Transformation which presence the magnitude of angle but not the sense are called isogonal.
( andθ φ have same magnitude and same sense)
Types of Transformation :
1. Translation : It is defined by w z= + α where α - is complex constant.
By this transformation each point in the z-plane is moved along the direction α through a
distance α by w z .= + α The figure in the z-plane are displaced in the direction of α keeping
their shape, size and orientation unchanged.
3. 3
- EXAMPLES –
Example 1 : Find the image of triangle with vertices 2, 1 i,1 i+ − under w z 1 i= − +
Solution : Let the vertices of the triangle in z-plane are demoted by 1 2 3z 2,z 1 i, z 1 i= = + = −
Here we have to find the vertices of triangle in w-plane and shape given transformation w z 1 i= − +
(translation type)
Now for z1 we have 1 1w z 1 i= − +
1 1w 2 1 i w 1 i⇒ = − + ⇒ = +
Similarly for z2 we have
2 2w z 1 i 1 i 1 i 2i= − + = + − + =
For 3 3z w O⇒ =
The image of triangle in z-plane is transformed to w-plane as shown in figure.
z-plane w-plane
Example 2 : Find the image of circle z 2= under the transformation w z 3 2i= + +
Solution : Given transformation
w z 3 2i= + +
z w (3 2i)⇒ = − +
z w (3 2i)⇒ = − + (taking mod)
2 w (3 2i)⇒ = − + ( z 2)=∵
There in the form of equation of circle where center in at 3 2i+ and radius 2. Thus equation of circle
z 2= in z-plane is transformed to w-plane as shown in figure
4. 4
2. Rotation :
Let the mapping w f(z)= be in the form i 0
w z.e θ
= where 0 realconstantθ =
By this transformation figure in the z-plane are rotated through an angle 0θ in w-plane.
Example 3 : With the help of transformation i /4
w ze π
= determine the region in w-plane corresponding to
the triangular region bounded by the lines x 0,y 0= = and x y 1+ = in z-plane
Solution : Given transformation i /4
w ze π
= (rotation type) put the value of w and z
u iv (x iy) cos isin
4 4
π π
+ = + +
1 1
(x iy) i
2 2
= + +
(By Euler formula i
e cos isinθ
= + θ )
1
(x iy)(1 i)
2
= + +
[ ]
1 1
(x xi iy y) (x y) i(x y)
2 2
= + + − = − + +
Equate real and Imaginary part on both sides
1
u (x y)
2
= − and
1
v (x y)
2
= + …………(1)
Now the region z-plane is bounded by the lines x 0,y 0, x y 1= = + =
For x 0= equation
1
(1) u ( y),
2
⇒ −
1
v (y) u v 0
2
= ⇒ + =
For y 0= equation
x x
(1) u ,v u v 0
22
⇒ = = ⇒ − =
For x y 1+ = equation
1
(1) v
2
⇒ =
Transform region in w-plane is bounded by the lines u v 0,+ = u v 0− = and
1
v
2
= Which in as shown
in figure.
5. 5
Example 4 : Find the image of the rectangular book by x 0,y 0,x 2= = = and y 3= under the
transformation w 2= i /4
e zπ
Solution : Given transformation w 2= i /4
e zπ
i /4
u iv 2. e (x iy)π
⇒ + = +
u iv 2 cos isin (x iy)
4 4
π π
⇒ + = + +
(by euler)
1 1
u iv 2 i (x iy)
2 2
⇒ + = + +
u iv (1 i)(x iy)⇒ + = + +
(x y) i(x y)= − + +
Equate the real and Imaginary part u x y,v x y= − = + Now the region in z-plane in bounded by line
x 0,= y 0,= x 2= and y 3=
For x 0= equation (1) u y,v y u v o⇒ = − = ⇒ + =
For y 0= equation (1) u x,v x,v x u v⇒ = − = = ⇒ =
For x 2= equation (1) u 2 y,v 2 y u v 4⇒ = − = + ⇒ + =
For y 3= equation (1) v x 3 i.e. u v 6⇒ = + − = −
Transformation region in w-plane is bounded by lines u v 0,+ = u v 0− = and u v 6− = − which is as
shown in figure.
Note : To plot the u v 4+ = we put
First u 0= we get v 4=
∴ Intercept in (0,4)
Similarly v 0 u 4= ⇒ =
∴ Intercept is (4,0)
u v 6− = −
3. Stretching or Magnification:
6. 6
This type of mapping is given by w az ,= a is real
4. Inversion :
This type mapping is defined by
1
w
z
=
It’s geometrical transformation is related to the inverse point with respect to the circle
Theorem 1 :
Under the transformation
1
w
z
= a straight line L in the z-plane is mapped into
a) a circle if does not pass through the origin z 0=
b) a straight line if L passes through the origin z 0=
Proof : Rewrite the mapping as
1
z
w
= i.e.
1
x iy
u iv
+ =
+
2 2
1 u iv u iv
x iy
u iv (u iv)(u iv) u v
− −
⇒ + = = =
+ + − +
2 2 2 2
u v
i
u v u v
= −
+ +
Equate real and Imaginary part on both sides
2 2 2 2
u v
x ,y
u v u v
= =
+ +
…………(1)
Any line L in the z-plane is given by the equation Ax By C 0+ + = where A, B, C are real for C 0,≠ it
does not pass through the origin …………(2)
2 2 2 2
u v
Equation (1) A B C 0
u v u v
∴ ⇒ + + =
+ +
2 2
C(u v ) Au Bv 0⇒ + + = = …………(3)
This represent a circle in the w-plane for C 0≠ hence the straight line L which does not pass
through the origin in mapped into a circle.
If L passes through the origin thenC 0= and then equation (3)⇒
Au Bv 0− =
This is straight line in the w-plane passes through the origin w 0=
Theorem 2 : The transformation
1
w
z
= always maps a circle into a circle with a straight line consider as
a circle of infinite radius.
7. 7
Proof : Now
1
w
z
= on solving we get
2 2
u
x ,
u v
=
+ 2 2
v
y
u v
= −
+
............(1)
Let 2 2
x y 2gx 2fy C 0+ + + + = ............(2)
Be any circle in the z-plane with C 0≠ It means that the circle does not pass through the origin
By equation (1) we get
2 2
2 2 2 2 2 2 2 2
u v u v
2g 2f C 0
u v u v u v u v
−
+ − + + + =
+ + + +
2 2
2 2 2 2 2 2 2
u v u v
2g 2f C 0
(u v ) u v u v
+
⇒ + + − + =
+ + +
2 2
C(u v ) 2gu 2 v 1 0⇒ + + − + + = …………(3)
Since C 0≠ , above represent a circle in the w-plane which also does not pass through the origin w 0.=
If C 0= then the circle equation (2) passes through z 0= and is mapped to equation (3) with C 0.=
2gu 2fu 1 0− + =
This is straight line in the w-plane considering a straight line as a circle of infinite radius above
equation represents a circle with infinite radius.
Hence every circle is mapped into circle under
1
w .
z
=
Example 5 : Under the transformation
1
w
z
= find the image of the circle z 2i 2− =
Solution : We know that
1
w
z
=
2 2 2 2
u v
x ,y
u v u v
⇒ = = −
+ +
…………(1)
The given circle is
z 2i 2− = i.e. x i(y 2) 2+ − =
2 2
x (y 2) 2⇒ + − =
2 2 2
x (y 2) 2⇒ + − = (squaring)
2 2
x y 4y 0⇒ + − = …………(2)
Under equation (1) equation (2) transform to
2 2
2 2 2 2 2 2
u v v
4 0
u v u v u v
−
+ − − =
+ + +
1 4v 0⇒ + = or
1
v
4
= −
This is a straight line parallel to the u-axis in the w-plane.
8. 8
Example 6 : Under the transformation
1
w
z
= find the image of the infinite strips
(a)
1 1
y
4 2
≤ ≤ and (b)
1
o y
2
< <
Solution : We know that from
1
w
z
= we get, 2 2
u
x ,
u v
=
+ 2 2
v
y
u v
= −
+
a) Under this equation the strip
1 1
y
4 2
≤ ≤ transform to 2 2
1 v 1
4 u v 2
−
≤ ≤
+
2 2
u v 4v 0⇒ + + ≤ and 2 2
u v 2v 0+ + ≥
2 2 2
(u 0) (v 2) 2⇒ − + + ≤ and 2 2 2
(u 0) (v 1) 1− + + ≥
Hence the transformed region in the w – plane consists of
1. Boundaries of the two circle
2 2 2
(u 0) (v 2) 2− + + ≥
This is circle with centre (0, 2)− and r 2= and 2 2 2
(u 0) (v 1) 1− + + = this is circle with center
(0, 1)− and radius 1
2. The area enclosed by the boundaries of these circle
The required region is shaded region
b) The region
1
0 y
2
< < is mapped into 2 2
v 1
0
u v 2
< − <
+
i.e. v 0< and 2 2 2
(u 0) (v 1) 1− + + > It
consists of the lower half of the w-plane without interior and the boundary of the circle
2 2 2
(u 0) (v 1) 1− + + = It is as shown in figure.
The Bilinear or Mobius Transformation :
The transformation
z
w ,
z
α + β
=
γ + δ
oαδ −βγ ≠ is called linear or fractional transformation where
, , andα β γ δ are complex constant.
9. 9
Theorem : A General linear transformation is a combination of the transformation of translation,
rotation and stretching.
Proof : Consider linear transformation
W z= α +β …………(1)
Where α and β are complex constant
Let i 0
a e 0θ
α = ≠ …………(2)
We define the following transformation
3f (z) azλ = = is magnification
i 0
2f ( ) e θ
ρ = λ = λ is rotating
1w f ( )= ρ = ρ +β is translation
Consider,
1 2 3 1 2 3(f f f )(z) f f [f (z)]=
1 2(f f )(az)=
1 2f [f (az)]=
i 0
1f (e az)θ
=
i 0
e az zθ
= +β =α +β [By equation (2)]
f(z)= [By equation (1)]
w=
Thus the L. T. is combination of translation rotation and magnification.
Theorem : The bilinear transformation is a combination of translation, rotation stretching and inversion.
Proof : The bilinear transformation is given by
z
w ,
z
α + β
=
γ + δ
oαδ −βγ ≠ …………(1)
Where , , andα β γ δ are complex constant
∴ equation (1) can be written as
(z / )
W
(z / )
α + β γ
=
γ + δ γ
Add /±δ γ
[(z / ) ( / / )]
[z / ]
α + δ γ + β α −δ γ
=
γ + δ γ
/ /
/ 1
(2 / )
β α −δ γ
=α γ +
+δ γ
W
z
µ
= λ +
+ η
…………(2)
10. 10
Where,
α
λ =
γ
( / / )α β α −δ γ
µ =
γ
/η =δ γ
We define
1 2 3 1 2 3f f f (z) f f [f (z)]=
1 2f f (z )= + η
1
1
f
z
=
+ η
z
µ
= + λ
+ η
W=
f(z)=
Thus the B.T. is combination of L.T., Inversion and translation. But L.T. is combination of
magnification, translation and rotation.
∴ B. T. in combination of translation, rotation magnification and inversion.
Fixed Point of Invariant Point :
The fixed or invariant point of the transformation w f(z)= are the points given by z f (z)=
Theorem : Every bilinear transformation with a single non-infinite fixed point can be put in the form.
1 1
k
W P Z P
= +
− −
(Proof in Book)
Theorem 2 : Every B. T. with two non-infinite fixed point P and Q is of the form
w P z P
k
w Q z Q
− −
=
− −
(Proof in Book)
Where ‘k’ is constant
Note : The B. T. having two fixed point P and Q is given by
w P z P
k
w Q z Q
− −
=
− −
where ‘k’ is constant then B. T. is
1. Parabolic : Fixed point one
2. Elliptic : Fixed point two, k 1=
3. Hyperbolic : Fixed point two, k 1≠ and
4. Loxodromic : Other than above
Example 7 : Find the fixed point and normal form of the B. T.
3z 4
w
z 1
−
=
−
11. 11
Solution : The give transformation is
3z 4
w
z 1
−
=
−
The fixed point by the transformation in is given by
3z 4
z
z 1
−
=
−
(by putting w z= )
2
z 4z 4 0⇒ − + =
2
(z 2) 0⇒ − =
z 2 P(say)⇒ = =
This has only on fixed point
∴ form of the transformation is
1 1
k
w P z P
= +
− −
1 1
3z 4w P P
z 1
=
−− −
−
1 1 z 1
3z 4w 2 3z 4 2z 22
z 1
−
⇒ = =
−− − − +−
−
z 1 z 2 1 1
1
z 2 z 2 z 2
− − +
= = = +
− − −
1 1
1
w 2 z 2
∴ = +
− −
Example 8 : Find the fixed point of B. T.
z
w
2 z
=
−
write its normal form state which is of type.
Solution : The given transformation is
z
w
2 z
=
−
…………(1)
The fixed point of the transformation (1) are given by
2z
z 2z z z
2 z
= ⇒ − =
−
2
z z 0⇒ − =
z(z 1) 0⇒ − =
z 0 or z 1⇒ = =
Let P 0= and Q 0= thus the given transformation has two fixed point. Thus the required transformation
must be of the form
w P z P
k
w Q z Q
− −
=
− −
Consider
w P z z
P Q
w Q 2 z 2 z
−
= − −
− − −
12. 12
z
0
w 0 2 z
zw 1 1
2 z
−
− −=
− −
−
Z 0 Z 0
2 z z 2z 2
− −
= =
− + + −
w 0 1 z 0
w 1 2 z 1
− −
⇒ =
− −
Here
1 1 1
K K
2 2 2
= ⇒ = =
Then the given transformation in hyperbolic
Ex. Find the fixed point of the transformation find the normal form and find its type.
Ex. Find the fixed point of the transformation
z 1
w
z 1
−
=
+
show that the transformation in elliptic
CIRCLE UNDER BILINEAR TRANSFORMATION
We have different from of equation representing a circle and they are as follows
C1 : Real equation of circle can be written as
2 2
x y 2gx 2fy C 0,+ + + + = g,f,C are real center ( g, f )= − − and radius 2 2
g f C= + −
C2 : Complex equation of circle are on follows
1. Z r,− α = α is complex, centre at z =α and radius = r
2. z 1= is unit circle
3. General equation zZ Bz Bz C 0+ + + = where ‘C’ is real center Z B= − and radius
BB C= −
4. 1
2
Z Z
k,
Z Z
−
=
−
K 0≥ this is family of equation of circle such that for k 1= it gives a straight line of
which 1 2z ,z are symmetric point.
Theorem : If 0z is in the upper half of z-plane then the bilinear transformation
i 0
0
z z
w e
z z
θ −
=
−
Maps the upper half of the z-plane into the interior of the unit circle in the w-plane
Proof : Proof in book.
Example 9 : Show that the map of real areas of z-plane onto w-plane by transformation
1
w
z i
=
+
is a
circle find its center and radius
13. 13
Solution : The real axis of z-plane is y 0= (z x iy)= +
If becomes z z 0− = …………(1)
z z 2iy
1
y (z z)
2i
− =
⇒ = −
∵
The given mapping
i.e.
1
w
z i
=
+
1
z i
w
⇒ + =
1
z i
w
⇒ = − + and
1
z i
w
= +
Then equation (1)
iw w w iw w w 0− + − − =
2i w w (w w) 0⇒ + − =
2 2
2i(u v ) 2iv 0⇒ + + =
w u iv
w u iv
w w 2iv
= +
= −
⇒ − =
2 2
u v v 0⇒ + + =
2 2
2 1 1
(u o) v
2 2
⇒ − + + =
This represent a circle in w-plane with center
1
o,
2
−
and radius
1
2
Example 10 : Prove that under the transformation
z i
w
iz 1
−
=
−
the region Imz o≥ is mapped into the
region w 1≤
Solution : For complex number z x iy= +
Imaginary of z i.e. Imz y=
Given transformation
z i
w
iz 1
−
=
−
Solving for Z we obtain
w i
z
iw 1
−
=
−
…………(1)
Given Imz 0 y 0≥ ⇒ ≥
i.e.
1
(z z) 0
2i
− ≥ then equation (1) ⇒
14. 14
1 w i w 1
0
2i iw 1 iw 1
− +
− ≥ − − −
ww 1
0
(ww 1)
−
⇒ ≥
− +
ww 1 0⇒ − ≤
ww 1⇒ ≤ ( ww 1 real 0)+ = >∵
Or w 1≤
Hence Imz 0≥ maps into the region w 1≤
Example 11 : Show that the transformation
2z 3
w
z 4
+
=
−
maps the circle 2 2
x y 4x 0+ − = into straight
line 4u 3 0+ =
Solution : As z x iy= + and z x iy= −
1
z z 2x x (z z)
2
∴ + = ⇒ = +
and 2 2i
y (z z), zz x y
2
= − − = +
Similarly for w u iv= + and w u iv= −
1
u (w w)
2
⇒ = +
and
i
v (w w)
2
= − − also 2 2
ww u v= +
putting this value in given equation of circle i.e. 2 2
x y 4x 0+ − =
we get z z 2(z z) 0− + = …………(1)
Now given transformation is
2z 3
w
z 4
+
=
−
Solving for Z we get
4w 3
z
w 2
+
=
−
Equation (1) ⇒
4w 3 4w 3 4w 3 4w 3
. 2 0
w 2 w 2 w 2 w 2
+ + + +
− + = − − − −
(4w 3)(4w 3) 2(4w 3)(w 2) 2(4w 3)(w 2) 0⇒ + + − + − − + − =
2(w w) 3 0⇒ + + =
2(2u) 3 0⇒ + = (asw w 2u)+ =
4u 3 0⇒ + =
4u 3 0+ =
This is equation of straight line thus circle is mapped onto the straight line 4u 3 0+ =
15. 15
Example 12 : Show that the transformation
5 4z
w
4z 2
−
=
−
transform the circle Z 1= into the circle of
radius unity in the w-plane.
Example 13 : Show that the relation
iz 2
w
4z i
+
=
+
transform the real axis in the z-plane into a circle in the
w-plane find the center and radius of circle.
Cross Ratio : If Z1, Z2, Z3, Z4 are district then the quantity
4 1 2 3
1 2 3 4
2 1 4 3
(Z Z )(Z Z )
(Z ,Z ,Z ,Z )
(Z Z )(Z Z )
− −
=
− −
…………(1)
is called the cross ratio of Z1, Z2, Z3, Z4
Easy way to Remember : Form the 4 cyclic difference of Z1, Z2, Z3, Z4
1 2Z Z− 2 3Z Z− 3 4Z Z− 4 1Z Z−
(1) (2) (3) (4)
Take (2) and (4) in numerator and (1) and (2) in denominator. This gives R.H.S. of equation (1)
Theorem : The cross Ratio remains invariant under a bilinear transformation
Proof : Consider a bilinear transformation
az d
w ,
cz d
+
=
+
ad bc 0− ≠
Assume that w1, w2, w3, w4 are the transform of Z1, Z2, Z3, Z4 respectively
then 4 1
4 1
4 1
az b az b
w w
cz d cz d
+ +
− = −
+ +
4 1 1 4
4 1
(az b)(cz d) (az b)(cz d)
(cz d) (cz d)
+ + − + +
=
+ +
4 1
4 1
z (ad bc) z (bc ad)
(cz d)(cz d)
− + −
=
+ +
4 1
4 1
4 1
(ad bc)(z z )
w w
(z d)(z d)
− −
⇒ − =
+ +
…………(1)
Similarly 2 3
2 3
4 1
(ad bc)(z z )
w w
(cz d)(cz d)
− −
− =
+ +
…………(2)
2 1
2 1
2 1
(ad bc)(z z )
w w
(cz d)(cz d)
− −
− =
+ +
…………(3)
and 4 3
4 3
4 3
(ad bc)(z z )
w w
(cz d)(cz d)
− −
− =
+ +
…………(4)
16. 16
4 1 2 3
1 2 3 4
2 3 4 3
(w w )(w w )
(w w w w )
(w w )(w w )
− −
∴ =
− −
4 1 2 3
2 1 4 3
(z z )(z z )
(z z )(z z )
− −
=
− −
1 2 3 4(z ,z ,z ,z )=
Thus the cross ratio is invariant under bilinear transformation
Example 14 : Find the bilinear transformation which maps the points z 1,= 1i 1− into the points
w 0,1,= ∞
Solution : Put the values of z and w in
az b
w
bz d
+
=
+
we get.
a b ai b a b
0 ,1 ,
C d Ci d C d
+ + − +
= = ∞ =
+ + − +
a b 0,Ci d ai b, C d 0⇒ + = + = + − + =
Solving these equation for a, c, d in terms of b we obtain
a b= − c d bi= = −
With these values the transformation
az b
w
Cz d
+
=
+
becomes
z 1
iz i
− +
=
− −
(We can try this by cross ratio method).
Example 15 : Find a bilinear transformation which maps points z o, i,= − − l into w i, , o= l respectively
Solution : Let the B.T. be
az b
w
cz d
+
=
+
…………(1)
For z 0,w i,= = equation
a(0) b b
(1) i
C(0) d d
+
⇒ = =
+
b id⇒ = …………(2)
17. 17
z i,w 1= − = equation
a( i) b
(1) i
C( i) d
− +
⇒ =
− +
…………(3)
z 1,w 0= − = , equation
a( 1) b
(1) o
(C 1) d
− +
⇒ =
− +
…………(4)
b a⇒ =
(2) and (4) id a⇒ = i.e. d ia= −
Then (3) C( i) ia ia a⇒ − − = − +
C ia⇒ =
Putting the values of b, c, d in terms of a in equations (1) we get
az a z 1 z 1
w i
ia z 1a i(Z 1) z 1
+ + +
= = = −
− − −
This is required transformation.
Objective Question :
1. A mapping that preserves angles between any oriented curves both in ………… and …………
is a conformal mapping.
2. Translation transformation is defined by the equation …………
3. Rotation transformation is defined by the equation …………
4. Stretching transformation is defined by the equation …………
5. Inversion transformation is defined by the equation …………
6. Bilinear Transformation given by …………
7. The fixed or invariant points of the transformation is given by …………
8. Normal form of B. T. with one fixed point say α is given by …………
9. Normal form of B. T. with two fixed point say P and Q is given by …………
10. The bilinear transformation with only one fixed point is called ………… transformation.
11. A bilinear transformation which is not parabolic, elliptic or hyperbolic is called …………
transformation.
*****