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NPTEL – Physics – Mathematical Physics - 1
Lecture 42
Jacobian of a transformation
Example 3. Jacobian of a transformation
For 𝜔 = 𝑓(𝑧) = √2𝑧𝑒
𝑖𝜋𝑖⁄4 + (1 − 2𝑖), find the Jacobian of the transformation.
Ans : The Jacobian is defined by,
𝐽 = 𝜕(𝑢,𝑣)
= | 𝜕
𝑥 𝜕(𝑥,𝑦)
𝜕𝑢 𝜕
𝑢
𝜕𝑦 𝜕
𝑥
𝜕𝑥 𝜕𝑦
𝜕𝑦
| = (𝜕𝑢
) + (𝜕𝑢
)
− 𝜕𝑢 𝜕𝑢
2 2
= |𝜕𝑢
+ 𝑖 𝜕𝑢
| = |𝑓′(𝑧)|2
𝜕𝑥 𝜕𝑦
2
Thus
2
𝐽 = |𝑓′(𝑧)|2 = |√2𝑒𝑖𝜋⁄4|
|𝑓′(𝑧)|2 is called the magnification factor
Example 4
A Möbious (or Moebius) transformation is any function of the form, (also known as
fractional linear or transformation or bilinear transformation)
𝜔 = 𝑓(𝑧) = 𝑎𝑧+
𝑏
𝑐𝑧+𝑑
with 𝑎𝑑 ≠ 𝑏𝑐 which would have made 𝜔 to be a constant function.
𝑓′(𝑧) = 𝑎𝑑−𝑏𝑐
2
𝑐𝑧+𝑑
𝑓′(𝑧) does not vanish, unless 𝑧 = − 𝑑⁄𝑐
The Mobius transformation 𝑓(𝑧) is conformal at every point except its pole 𝑧 = − 𝑑⁄𝑐
Example 5
Find the image of the interior of the circle : |𝑧 − 2| = 2 under the Mobius
transformation
𝜔 = 𝑓(𝑧) =
𝑧
2𝑧−8
Ans : Since f has a pole at 𝑧 = 4 and this point lies on 𝐶, the image has to be a straight
line connecting
Joint initiative of IITs and IISc – Funded by MHRD Page 61 of 66

𝜔 = 𝑓(0) = 0 and 𝜔 = 𝑓(𝑧 + 𝑧𝑖) = −𝑖
2
𝜔 = 𝑓(𝑧) = − 1
2
Geometrically,
Example 6
Find the conformal map of unit circle centered at origin |𝑧| = 1 onto the right half plane
plane 𝑅𝑒 𝜔 > 0
Ans: We have to look for a (bilinear) transformation that maps the circle |𝑧| = 1 onto the
imaginary axis of the complex plane. The transformation must therefore have a pole on
the circle. Further, the origin 𝜔 = 0 must also lie on the image of the circle.
So consider 𝜔 = 𝑓1(𝑧) = 𝑧−1
where 𝜔 maps -1 to 0 and 1 to ∞. Thus 𝑓1(𝑧) maps |𝑧| = 1
onto some straight line through the origin.
𝑧+1
To find out which straight line, we plug in z = 1 and find that the point
𝑖−1
𝜔 = 𝑖+1
= −𝑖
Which also lies on the same line.
Thus the image of the circle |𝑧| = 1 under 𝑓1(𝑧) must be the imaginary axis.
To obtain which half plane is the image of the interior of the circle, we check the point z
= 0. It is mapped by 𝑓1(𝑧) to the point 𝜔 = −1 in the left half phase. However since we
want the right half plane, a rotation of 𝜋 can be applied to yield,

𝜔 = 𝑓(𝑧) = − 𝑧+1
= 1+𝑧
𝑧−1 1−𝑧
Any further vertical translation is also allowed. Thus geometrically we can represent,
Tutorial Problems
1.Find a bilinear transformation that maps the region 𝐷1: |𝑧| > 1 onto a region
𝐷2: 𝑅𝑒𝜔 < 0.
2.Find a transformation that maps the interior of the circle |𝑧 − 𝑖| = 2
onto the exterior of the circle |𝜔 − 𝑖| = 3
3.Find a linear fractional transformation that maps the half plane defined by 𝐼𝑚(𝑧) >
𝑅𝑒(𝑧) onto the interior of the circle, |𝜔 − 𝑖| = 3
4.Prove that the Jacobian of a transformation is unitary, that is,
𝜕(𝑥,𝑦) 𝜕(𝑢,𝑣)
=
1
𝜕(𝑢,𝑣) 𝜕(𝑥,𝑦
)

Physical applications of Conformal mapping
A number of problems in science and Engineering can be conveniently solved
using conformal mapping. Especially we shall show the applications on harmonic
and conjugate functions.
A function Φ that satisfies Laplace’s equation,
𝜕𝑥2 𝜕𝑦 2
∇2Φ = 𝜕 Φ
+ 𝜕 Φ
= 0
2 2
In a region 𝑅′ is called harmonic in R. As 𝑓(𝑧) = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) is analytic in R, then
𝑢 and 𝑣 are harmonic in 𝑅 .
A brief introduction to Dirichlet and Neumann problems will be helpful. Let R be a
simply connected region bounded by a simple closed curve 𝐶 as shown in the figure.
a) Dirichlet problem
The function that satisfies Laplace’s equation in 𝑅 and assumes prescribed values on the
boundary C.
b) Neumann problem
The function that satisfies Laplace’s equation in R and assumes prescribed values
𝜕Φ
for the normal derivatives on the boundary C.
𝜕𝑛
Heat Conduction – Steady temperature in a region
The Fourier’s law of heat conduction is stated as,
Φ = −𝑘 𝑑𝑇
𝑑𝑁
(𝑘 > 0) (1)
Where Φ is a heat flux flowing normal to a surface per unit area per unit time, 𝑘
is a constant and is known as the thermal conductivity of the material, 𝑇 is the
temperature

function and is the normal derivative of the temperature 𝑇 at a point on the surface. It
𝑑𝑇
𝑑𝑁
is known that (not proved here) that the temperature distribution satisfies Laplace’s
equation at each point inside the material, namely,
𝑇𝑥𝑥 (𝑥, 𝑦) + 𝑇𝑦𝑦 (𝑥, 𝑦) =
0
(2)
Thus, coupled with continuity of the temperature function and its partial derivatives, 𝑇 is
a harmonic function in a domain that denotes the interior of the material under
consideration. The surfaces 𝑇(𝑥, 𝑦) = constant are called as isotherms which are
level curves of the function T.
Let us consider a typical problem of solving a boundary value problem in heat
conduction using conformal mapping. It may be reminded to the readers that the same
problem may have been solved by using the method of separation of variable which
would yield the solution in the form of an infinite series.
A steady state temperature distribution is to be found for the region shown as below
The boundary condition is such that 𝑇 = 100°𝑐 along the y- axis (𝑥 = 0) and
𝑇 = 50°𝑐 along the wall of a cylinder whose cross sectional area on the 𝑥𝑦 plane
is denoted by
|𝑧 − 1
|
2 2
= 1
One can choose the transformation from 𝑥𝑦 to 𝑢𝑣 plane (a from z to w plane) as
𝑤 = 1
𝑧
In order to transform the shaded region in the 𝑥𝑦 plane, we note that

(i) 𝑥 = 0 (y axis) is equivalent to
𝑧 + 𝑧̅ = 0
𝑤 𝑤 𝑤𝑤̅
Thus 𝑧 + 𝑧̅ = 1
+ 1
= 𝑤+𝑤̅
= 0
So 𝑧 + 𝑧̅ = 0 is mapped onto 𝑤 + 𝑤̅ = 0, which imples y-axis is mapped onto v-axis
(ii) |𝑧 − 1
| = 1
(circle centered at 1
with a radius 1
as shown in the figure)
2 2 2 2
Substituting 𝑧 = 1
yields
𝑤
| 1
− 1
| = 1
𝑤 2 2
Or, |𝑤 − 2| = |𝑤 − 0|
Which represents the middle line between 𝑤 = 2 and 𝑤 = 0, which is 𝑢 = 1
So the shaded domain in the 𝑥𝑦 plane is mapped onto an infinite ship surrounded
by
𝑢 = 0 and 𝑢 = 1 . Along the 𝑣 axis, the mapped region extends to ±∞. Thus the
temperature distribution, T will have no dependency on y. Hence
𝑇(𝑢, 𝑣) = −50𝑥 + 100
Thus 𝑇(𝑥, 𝑦) =
50𝑥
𝑥2+𝑦 2 + 100
Tutorial
Using the conformal mapping 𝑤 = 𝑧2, find the temperature distribution in the shaded
region.

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