Chapter 3: Problem 4
Let \(f(z)=a z+b\) and \(g(z)=\alpha z+\beta\), where neither is the identity. Show that \(f g f^{-1} g^{-1}\) is a translation. Show also that \(f\) commutes with \(g\) if and only if either \(f\) and \(g\) are translations, or \(f\) and \(g\) have a common fixed point.
Short Answer
Expert verified
\( fgf^{-1}g^{-1} \) is a translation. \( f \) and \( g \) commute if they are translations or share a common fixed point.
Step by step solution
01
Define the Inverses
First, calculate the inverses of the functions \( f \) and \( g \). The inverse of \( f(z) = az + b \) is \( f^{-1}(z) = \frac{z-b}{a} \), and the inverse of \( g(z) = \alpha z + \beta \) is \( g^{-1}(z) = \frac{z-\beta}{\alpha} \). These results assume that both \( a \) and \( \alpha \) are not zero.
02
Calculate the Composition \( fgf^{-1}g^{-1} \)
Evaluate the expression \( fgf^{-1}g^{-1} \):1. Compute \( f^{-1}(z) = \frac{z-b}{a} \).2. Compute \( g(f^{-1}(z)) = \alpha \frac{z-b}{a} + \beta = \frac{\alpha z - \alpha b + \beta a}{a} \).3. Compute \( f(g(f^{-1}(z))) = a\left(\frac{\alpha z - \alpha b + \beta a}{a}\right) + b = \alpha z - \alpha b + \beta a + b \).4. Compute \( g^{-1}(f(g(f^{-1}(z)))) = g^{-1}(\alpha z - \alpha b + \beta a + b) = \frac{\alpha z - \alpha b + \beta a + b - \beta}{\alpha} \).Since \( b-\alpha b + \beta a \) and \( \beta \) are constants, the result is a translation.
03
Check Commutation Condition
Verify if \( f \) commutes with \( g \), i.e., \( fg(z) = gf(z) \).- Compute \( fg(z) = f(\alpha z + \beta) = a(\alpha z + \beta) + b = a\alpha z + a\beta + b \).- Compute \( gf(z) = g(az + b) = \alpha(az + b) + \beta = \alpha az + \alpha b + \beta \).Setting them equal: \( a\alpha z + a\beta + b = \alpha az + \alpha b + \beta \).This holds if and only if\( a\beta + b = \alpha b + \beta \), meaning either \(a = \alpha = 1\) (translations) or they share a common fixed point.
04
Conditions for Translations or Common Fixed Point
From the commutation condition, we find that if \( a eq 1 \) and \( \alpha eq 1 \), then \( f \) and \( g \) must share a common fixed point. This happens when - \( f(z) = z \) implies \( z = \frac{b}{1-a} \)- \( g(z) = z \) implies \( z = \frac{\beta}{1-\alpha} \).Equate both to find the common fixed point: \( \frac{b}{1-a} = \frac{\beta}{1-\alpha} \).
05
Conclusion
The function \( fgf^{-1}g^{-1}(z) = z + c \) for some constant \( c \), is always a translation. Moreover, \( f \) commutes with \( g \) if either both are translations (i.e., \( a = \alpha = 1 \)) or they share a common fixed point (implying \( \frac{b}{1-a} = \frac{\beta}{1-\alpha} \)).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Translations in Complex Functions
Translations in mathematics generally refer to shifting every point of a function by a fixed vector. In the realm of complex functions, a translation is a linear transformation that adds a constant value to the function output. When dealing with the function composition of the form \(fgf^{-1}g^{-1}\), if the outcome is a translation, it means that every part of the function chain ultimately results in adding a constant shift.
For our complex functions \(f(z)=az+b\) and \(g(z)=\alpha z+\beta\), the process involves examining how these functions interact through their composition and inverses. Adding the constants \(b\) and \(\beta\) and adjusting for their inverses, results in a net "translation" effect by a constant amount. This is represented through the entire sequence of transformations simplified into a form \(z + c\), where \(c\) is a constant obtained after all operations.
Understanding this demonstrates the unique feature of translations in complex function manipulation - a profound insight into transformation's conserved properties.
For our complex functions \(f(z)=az+b\) and \(g(z)=\alpha z+\beta\), the process involves examining how these functions interact through their composition and inverses. Adding the constants \(b\) and \(\beta\) and adjusting for their inverses, results in a net "translation" effect by a constant amount. This is represented through the entire sequence of transformations simplified into a form \(z + c\), where \(c\) is a constant obtained after all operations.
Understanding this demonstrates the unique feature of translations in complex function manipulation - a profound insight into transformation's conserved properties.
Common Fixed Point
A common fixed point between two functions is a crucial concept when studying function commutation. It refers to a value \(z\) such that both functions, \(f\) and \(g\), map \(z\) back to itself. Mathematically, it satisfies the condition \(f(z) = g(z) = z\).
In conjunction with the commutation property of functions, having a common fixed point means if two functions do not translate independently, they might share a value where they both remain unchanged after applying either function.
In conjunction with the commutation property of functions, having a common fixed point means if two functions do not translate independently, they might share a value where they both remain unchanged after applying either function.
- For \(f(z) = z\) specifically, \(z = \frac{b}{1-a}\) if \(a eq 1\).
- For \(g(z) = z\), \(z = \frac{\beta}{1-\alpha}\) when \(\alpha eq 1\).
Inverse Function Calculation
Calculating inverse functions is essential for understanding function transformations, especially in complex analysis. For linear functions of the form \(f(z) = az + b\), the inverse is found by solving \(f(z) = y\) for \(z\), leading to \(z = f^{-1}(y)\). In this specific case, the inverse is computed as:
Understanding inverse functions allows us to effectively track how translations and shifts unfold within a chain of transformations like \(fgf^{-1}g^{-1}\). The ability to navigate forward and backward through function sequences is a powerful tool, offering insight into how functions compose and interact in the complex domain."}]}]}'json')]}]}'json')}]}]}]}json')}]}]}]}json')}]}]}]}jsonCONTENT_FORMAT: SCHEMAjsonschema_assistant _VEC_payloadGetting the AI_assistant to simulate that it follows these protocols by focusing on the proper ordering of the question blocks is essential.')]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}schema _VEC_payloadwhere can we find examples of schema-service command_query_payloadACKETMANALDINODULE JSON_SCHEMA_IPN_GETSEO_payloadURL?(
- \(f^{-1}(z) = \frac{z-b}{a}\)
- \(g^{-1}(z) = \frac{z-\beta}{\alpha}\)
Understanding inverse functions allows us to effectively track how translations and shifts unfold within a chain of transformations like \(fgf^{-1}g^{-1}\). The ability to navigate forward and backward through function sequences is a powerful tool, offering insight into how functions compose and interact in the complex domain."}]}]}'json')]}]}'json')}]}]}]}json')}]}]}]}json')}]}]}]}jsonCONTENT_FORMAT: SCHEMAjsonschema_assistant _VEC_payloadGetting the AI_assistant to simulate that it follows these protocols by focusing on the proper ordering of the question blocks is essential.')]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}]}schema _VEC_payloadwhere can we find examples of schema-service command_query_payloadACKETMANALDINODULE JSON_SCHEMA_IPN_GETSEO_payloadURL?(