Problem 11
Question
\(S(w)=\frac{\left(w-w_{1}\right)\left(w_{2}-w_{3}\right)}{\left(w-w_{3}\right)\left(w_{2}-w_{1}\right)}\) maps \(w_{1}, w_{2}, w_{3}\) to \(0,1, \infty\) and so \(S^{\prime}\) maps \(0,1, \infty\) to \(w_{1}, w_{2}, w_{3} .\) Therefore, \(z=\frac{(w-0)(i-2)}{(w-2)(i-0)}\) and so \(w=\frac{2 z}{z-1-2 i}\) maps \(0,1, \infty\) to \(0, i, 2\).
Step-by-Step Solution
Verified Answer
The function correctly maps 0, 1, ∞ to 0, i, 2.
1Step 1: Understand the Mapping Function
The function \( S(w) = \frac{(w - w_1)(w_2 - w_3)}{(w - w_3)(w_2 - w_1)} \) is given as a mapping from the points \( w_1, w_2, w_3 \) to \( 0, 1, \infty \), respectively. This is a Möbius transformation or a linear fractional transformation.
2Step 2: Find the Inverse Mapping
We need to find the function \( S' \) that performs the inverse mapping from \( 0, 1, \infty \) back to \( w_1, w_2, w_3 \). Given the transformation properties, recognize that the transformed coordinates make the inverse a direct algebraic manipulation of the original function.
3Step 3: Analyze Given Conditions
From the statement, the transformation \( z=\frac{(w-0)(i-2)}{(w-2)(i-0)} \) is provided, mapping certain values to new points. Ensure understanding by setting \( z \) values as \( 0, 1, \infty \) and verifying the mapped values are consistent.
4Step 4: Derive Expression for \( w \)
Given \( z=\frac{(w-0)(i-2)}{(w-2)(i-0)} \), solve for \( w \) to express it in terms of \( z \). Simplify the equation using algebraic manipulation to obtain \( w = \frac{2z}{z-1-2i} \).
5Step 5: Verify the Map
Check that the derived expression for \( w \) correctly maps \( z = 0, 1, \infty \) onto the desired points \( 0, i, 2 \). Substitute \( z \) values into \( w = \frac{2z}{z-1-2i} \) and check if the results match \( 0, i, 2 \).
Key Concepts
Complex MappingInverse FunctionLinear Fractional TransformationAlgebraic Manipulation
Complex Mapping
Complex mapping is an essential concept in complex analysis, dealing with functions that map complex numbers from one form or domain to another. In the given exercise, the function \( S(w) = \frac{(w - w_1)(w_2 - w_3)}{(w - w_3)(w_2 - w_1)} \) represents a specific kind of mapping known as a Möbius transformation or linear fractional transformation.
It transforms the complex points \( w_1, w_2, \text{ and } w_3 \) into the canonical points \( 0, 1, \text{ and } \infty \).
This specific mapping is significant because it reshapes the complex plane in a way that conserves cross-ratios, a fundamental property in projective geometry. Understanding this transformation involves recognizing how the function alters the original complex points and understanding the properties that remain intact, like cross-ratio, under such a transformation.
It transforms the complex points \( w_1, w_2, \text{ and } w_3 \) into the canonical points \( 0, 1, \text{ and } \infty \).
This specific mapping is significant because it reshapes the complex plane in a way that conserves cross-ratios, a fundamental property in projective geometry. Understanding this transformation involves recognizing how the function alters the original complex points and understanding the properties that remain intact, like cross-ratio, under such a transformation.
Inverse Function
Finding the inverse of a function is about determining a new function that reverses the effect of the original transformation. For the Möbius transformation given as \( S(w) = \frac{(w - w_1)(w_2 - w_3)}{(w - w_3)(w_2 - w_1)} \), the inverse mapping \( S' \) maps \( 0, 1, \infty \) back to \( w_1, w_2, w_3 \).
The inverse function undoes the mapping, recreating the original set of points in the complex plane. Finding this inverse involves a series of steps where you swap and rearrange terms algebraically. This often involves flipping terms in the numerator and denominator, followed by careful re-substitution of the coordinates that the function maps.
Recognizing how each point inversely maps ensures a symmetrical and perfectly reversible transformation.
The inverse function undoes the mapping, recreating the original set of points in the complex plane. Finding this inverse involves a series of steps where you swap and rearrange terms algebraically. This often involves flipping terms in the numerator and denominator, followed by careful re-substitution of the coordinates that the function maps.
Recognizing how each point inversely maps ensures a symmetrical and perfectly reversible transformation.
Linear Fractional Transformation
Linear fractional transformations, also known as Möbius transformations, are special transformations in complex analysis that can be represented as \( S(z) = \frac{az + b}{cz + d} \), where \( a, b, c, \text{ and } d \) are complex numbers and \( ad - bc eq 0 \).
This particular transformation is notable for its ability to map circles and lines to other circles and lines within the complex plane.
Möbius transformations are bijective on the Riemann sphere, which includes the point at infinity, ensuring uniqueness in mapping complex points from one form to another.
This allows transformations of the simplest sequences of points like \( 0, 1, \infty \) to any trio of points like \( w_1, w_2, w_3 \), making these transformations extremely useful in conformal mappings and complex analysis.
This particular transformation is notable for its ability to map circles and lines to other circles and lines within the complex plane.
Möbius transformations are bijective on the Riemann sphere, which includes the point at infinity, ensuring uniqueness in mapping complex points from one form to another.
This allows transformations of the simplest sequences of points like \( 0, 1, \infty \) to any trio of points like \( w_1, w_2, w_3 \), making these transformations extremely useful in conformal mappings and complex analysis.
Algebraic Manipulation
Algebraic manipulation involves altering expressions to simplify or change their form, often to find solutions or new representations. In the context of the exercise, algebraic manipulation is key to deriving the expression for \( w \) given \( z \) by solving \( z=\frac{(w-0)(i-2)}{(w-2)(i-0)} \).
To arrive at the simplified form \( w = \frac{2z}{z-1-2i} \), each step involves carefully rearranging the terms to isolate \( w \).
This manipulation is crucial in complex function theory as it enables solving equations by balancing components across the equation.
To arrive at the simplified form \( w = \frac{2z}{z-1-2i} \), each step involves carefully rearranging the terms to isolate \( w \).
This manipulation is crucial in complex function theory as it enables solving equations by balancing components across the equation.
- Start by expanding terms using distributive properties.
- Reorganize terms for easier isolation.
- Utilize cross-multiplying to deal with fractions.
- Simplify the expression by factoring or canceling common terms.
Other exercises in this chapter
Problem 10
If \(w=z+\frac{1}{z}\) and \(z=e^{i t}, w=e^{i t}+e^{-i t}=2 \cos t .\) Therefore \(u=2 \cos t\) and \(v=0\) and so the image is the closed interval [-2,2] on t
View solution Problem 11
(a) \(\psi(x, y)=\operatorname{Im}(\sin z)=\cos x \sinh y\) and \(\psi(x, y)=0\) when \(x=\pm \pi / 2\) or when \(y=0\) (b) \(\mathbf{V}=\overline{G^{\prime}(z)
View solution Problem 11
The first quadrant may be described by \(r>0,0
View solution Problem 14
Using the cross-ratio formula (7), $$\frac{(1)(-i-1)}{(w-1)(1)}=\frac{(z+1)(-1)}{(z-1)(1)}$$ and so \(w=\frac{(2+i) z-i}{z+1}\) maps -1,0,1 to \(\infty,-i, 1\).
View solution