Conformal Mapping

Conformal mapping is an important concept in complex analysis that refers to a function that preserves angles and shapes of infinitesimally small figures, though it may change their size. It is a type of transformation that preserves angles between every pair of curves through a point, both in magnitude and sense, and is said to be conformal at that point.

A mapping f: C→C is said to be conformal at a point z0 if:

1. f is analytic (differentiable) at z₀​, and
2. The derivative f′(z0) ≠ 0.

• The function f preserves the angle between two intersecting curves at the image point; that is, if two curves meet at an angle θ at z₀​, then their images under f also meet at the same angle θ.
• The magnitude of f′(z₀) represents the scaling (how much the size changes).
• The argument of f′(z₀) represents the rotation applied at that point.

So locally, f acts like:

f(z) ≈ f(z₀​) + f′(z₀​) (z − z₀​)

which is just a combination of scaling and rotation.

Graphical Representation of Conformal Mapping

Consider the transformations m = m(h, k) and n = n(h, k), which map a point R(h₀, k₀) in the α-plane to a point R′(m₀, n₀) in the β-plane. Let curves E₁ and E₂ intersect at point α₀ = (h₀, k₀) in the α-plane and be mapped into curves E₁′ and E₂′ in the β-plane intersecting at β₀ = (m₀, n₀).

If the transformation is such that the angle between E₁ and E₂ at α₀ is equal both in magnitude and sense to the angle between the curves E₁′ and E₂′ at β₀, it is known as a conformal mapping at α₀ = (h₀, k₀).

Conditionals for Conformal Mapping

A function f(z) is called conformal at a point if it preserves angles and their orientation at that point. To ensure this, the function must satisfy the following mathematical conditions:

Analyticity at the Point z₀

For a mapping to behave smoothly and allow angle preservation, the function must first be analytic. This requires that the complex derivative at z₀ exists:

f′(z_0)=lim_{z→ 0}{\frac{f(z)−f(z_0)}{z−z_0}}

Non-zero Derivative at z₀

Even if the function is analytic, it will fail to be conformal if the derivative becomes zero. Thus, the second condition is:

f′(z₀ ) ≠ 0

Conformal on a Region

A function f(z) is conformal on a region R if:

• It is analytic on the entire region, and
• Its derivative does not vanish anywhere in R.

Solved Questions on Conformal Mapping

Question 1: Show that the map w=\frac{2}{\sqrt z}-1

transforms the region outside the parabola y² = 4(1−x) in the z-plane into the interior of the unit circle ∣w∣ < 1 in the w-plane.

Solution :

Start from the map and rearrange:

w=\frac{2}{\sqrt z}-1 , w+1 =\frac{2}{\sqrt z}

Square both sides (careful with branches; we will use polar form later):

(w+1)² = 4/z or z =\frac{4}{(w+1)^{2}}.

Boundary: take ∣w∣ = 1. Put w = e^iϕ. Then by (1)

z = 4/(e^iϕ+1)².

Write e^iϕ+1 = 1 + cos⁡ϕ + isin⁡ϕ. It is convenient to factor e^iϕ/2:

e^iϕ+1 = e^iϕ/2(e^−iϕ/2+e^iϕ/2) = 2e^iϕ/2cos⁡ϕ/2

z = \frac{(e^{i\phi} + 1)^2}{4}= \cos^2\left(\frac{\phi}{2}\right)e^{-i\phi}

x = \cos^2\left(\frac{\phi}{2}\right)\cos\phi, \qquad y = -\cos^2\left(\frac{\phi}{2}\right)\sin\phi

Using identities:

x = 1 − t²,  y = −2t (t = tan⁡(ϕ/2)) x = 1 - t²

Eliminate t: y² = 4(1−x)y²

Hence, ∣w∣ < 1 corresponds to the exterior of the parabola y² = 4(1−x)y².

Question 2: Determine the region ‘D′’ of the w-plane into which the triangular region D enclosed by the lines x = 0, y = 0, and x + y = 1 in the z-plane is transformed under the transformation: w = z².

Solution:

Let

z = x + iy

Then

w = z² = (x + iy)² = (x² − y²) + i(2xy)

So,

u = x² − y², v = 2xy

where w = u + iv
The region is bounded by:

x = 0,y = 0, x + y = 1

It is a right-angled triangle with vertices:

A(0, 0), B(1, 0), C(0, 1)

Line y = 0:
For y = 0,

u = x²,v = 0

As x goes from 0 to 1,

u = 0 to 1,v = 0

This represents the segment of the real axis from w = 0 to w = 1.

For x = 0,

u = −y², v = 0

As y goes from 0 to 1,

u = 0 to −1, v = 0

This represents the segment of the real axis from w = 0 to w = −1 .

Let y = 1 − x.
Then,

u = x²−(1 − x)²= 2x − 1

2x(1 −x) = 2x−2x²

For x = 0, (u, v) = (−1,0)(u, v)
For x = 1, (u,v) = (1,0)

So, as x varies from 0 to 1, the curve is

u = 2x−1, v = 2x − 2x²

Eliminating x:

v = 1 − u²

which represents a parabola in the w-plane, opening downward with vertex at (0, 1).

The image of the triangular region in the z-plane is the region bounded by:

u = −1 to 1,  v = 0,  and v = 1 − u²

Hence,
the region D′ in the w-plane is the area bounded by the parabola v = 1−u² and the real axis between u = −1 and u = 1.

The transformation w = z² maps the triangular region bounded by x = 0,y = 0,x + y = 1 in the z-plane into the region bounded by the parabola v = 1−u² and the u-axis from u = − 1in the w-plane.

Unsolved Question on Conformal Mapping

Question 1: Determine the region D′ of the w-plane into which the triangular region D enclosed by the lines x = 0,y = 0,x + y = 1 in the z-plane is transformed under the transformation w = 1 + 2z.

Question 2: Show that the transformation w = 1/z maps the region outside the circle : x² + y² = 1 in the z-plane into the interior of the circle u² + v² = 1 in the w-plane.

Question 3: Show that the transformation w = z² maps the upper half of the z-plane (y > 0) onto the entire w-plane cut along the negative real axis.

Question 4: Let D be the triangular region in the z-plane bounded by x = 0,y = 0,x + y = 1 (with vertices at 0, 1, i). Determine the image region D′ in the w-plane under the Möbius transformation w = z − 1/z + 1