Problem 5
Question
The mapping \(f(z)=z^{4}\) maps the wedge \(0 \leq \operatorname{Arg} z \leq \pi / 4\) to the upper half-plane \(R^{\prime}\) and \(U=\frac{1}{\pi}\) Arg \(w\) is the solution to the corresponding Dirichlet problem in \(R^{\prime} .\) Therefore, \\[ \phi(x, y)=U\left(z^{4}\right)=\frac{1}{\pi} \operatorname{Arg} z^{4}=\frac{4}{\pi} \operatorname{Arg} z \\] is the potential in the wedge. A complex potential is \(G(z)=\frac{4}{\pi} \mathrm{Ln} z\) and, since \(\phi(r, \theta)=\frac{4}{\pi} \theta,\) the equipotential lines are the rays \(\theta=\frac{\pi}{4} c .\) Finally \\[ \mathbf{F}=\overline{G^{\prime}(z)}=\frac{4}{\pi} \frac{1}{z}=\frac{4}{\pi}\left(\frac{x}{x^{2}+y^{2}}, \frac{y}{x^{2}+y^{2}}\right) \\]
Step-by-Step Solution
Verified Answer
The equipotential lines are \( \theta = \frac{\pi}{4} c \), and the field is \( \mathbf{F} = \frac{4}{\pi} \left(\frac{x}{x^2 + y^2}, \frac{y}{x^2 + y^2}\right) \).
1Step 1: Understand the Mapping and Wedge Region
The given function is \( f(z) = z^4 \), which maps the complex number \( z \) to another in the form of raising \( z \) to the fourth power. The specified wedge is the region in the complex plane where the argument of \( z \), \( \text{Arg}(z) \), is between \( 0 \) and \( \pi/4 \). This region maps onto the upper half-plane denoted as \( R' \).
2Step 2: Determine the Potential Function
The potential function in the wedge \( \phi(x, y) = U(z^4) \) is given by \( \phi(x, y) = \frac{1}{\pi} \text{Arg}(z^4) = \frac{4}{\pi} \text{Arg}(z) \). This is because \( z^4 \) multiplies the argument of \( z \) by 4.
3Step 3: Find the Complex Potential
The complex potential related to the problem is given by \( G(z) = \frac{4}{\pi} \text{Ln}(z) \). This is derived from the relationship between the potential \( \phi \) and the argument of \( z \).
4Step 4: Equipotential Lines
The equipotential lines in this context are defined by \( \phi(r, \theta) = \frac{4}{\pi} \theta \), which leads to the expression \( \theta = \frac{\pi}{4} c \). These are rays emerging from the origin at angles related to a constant \( c \).
5Step 5: Calculate the Field \( \mathbf{F} \)
The field \( \mathbf{F} \) related to the potential is found by taking the complex conjugate of the derivative of the complex potential: \( G'(z) \). We have \( G'(z) = \frac{4}{\pi} \frac{1}{z} \), and thus \( \mathbf{F} = \overline{G'(z)} = \frac{4}{\pi} \left(\frac{x}{x^2 + y^2}, \frac{y}{x^2 + y^2}\right) \). This represents the field in Cartesian coordinates.
Key Concepts
Dirichlet ProblemComplex MappingComplex PotentialEquipotential Lines
Dirichlet Problem
The Dirichlet Problem is a fundamental question in the field of complex analysis. It involves finding a function that satisfies a specific condition on the boundary of a given domain. In this particular problem, we are looking at a scenario in the complex plane. The function should be harmonic, meaning there are no singularities within the domain.
In our exercise, we use the function \( U = \frac{1}{\pi} \operatorname{Arg} w \) as our solution to the Dirichlet Problem. The original domain, known as the wedge, is mapped onto the upper half-plane. Here, the solution \( U \) ensures that boundary values are maintained as required by the problem's conditions. The manipulation of angles through argument functions helps solve the boundary constraints, preserving the harmonic nature of the function.
In our exercise, we use the function \( U = \frac{1}{\pi} \operatorname{Arg} w \) as our solution to the Dirichlet Problem. The original domain, known as the wedge, is mapped onto the upper half-plane. Here, the solution \( U \) ensures that boundary values are maintained as required by the problem's conditions. The manipulation of angles through argument functions helps solve the boundary constraints, preserving the harmonic nature of the function.
Complex Mapping
Complex Mapping, in this context, refers to the transformation of complex numbers according to the function \( f(z) = z^4 \). This mapping takes a complex number \( z \) and raises it to the power of four, significantly altering its position in the complex plane.
When we apply this transformation, we map a wedge-like region, defined by \( 0 \leq \operatorname{Arg} z \leq \frac{\pi}{4} \), onto the upper half-plane. The mapping essentially spins the region in such a manner that every point's argument is multiplied by four, effectively enlarging and rotating the original angle domain. This forms the basis for solving the Dirichlet Problem given in this exercise, shifting our focus from purely geometric reasoning to algebraic manipulation with complex numbers.
When we apply this transformation, we map a wedge-like region, defined by \( 0 \leq \operatorname{Arg} z \leq \frac{\pi}{4} \), onto the upper half-plane. The mapping essentially spins the region in such a manner that every point's argument is multiplied by four, effectively enlarging and rotating the original angle domain. This forms the basis for solving the Dirichlet Problem given in this exercise, shifting our focus from purely geometric reasoning to algebraic manipulation with complex numbers.
Complex Potential
Complex Potential encompasses functions that describe potential flows in the complex plane. Here, it is closely related to our potential function \( \phi(x, y) \), often used in physics and engineering to represent fields or energy levels.
The complex potential for this problem is expressed as \( G(z) = \frac{4}{\pi} \ln z \). This function stems from the integral of the potential, which is linked to the argument of \( z \). By focusing on \( G(z) \), we can express our potential in an analytic form, aiding in the computation of various field properties. The logarithmic function naturally connects the potential difference with the location in the complex plane, providing a powerful tool for exploration and application.
The complex potential for this problem is expressed as \( G(z) = \frac{4}{\pi} \ln z \). This function stems from the integral of the potential, which is linked to the argument of \( z \). By focusing on \( G(z) \), we can express our potential in an analytic form, aiding in the computation of various field properties. The logarithmic function naturally connects the potential difference with the location in the complex plane, providing a powerful tool for exploration and application.
Equipotential Lines
Equipotential Lines represent locations where the potential function remains constant, crucial in visualization techniques. Within this context, these lines translate to distinct rays emanating from the origin.
Here, the equipotential lines are derived from the expression \( \phi(r, \theta) = \frac{4}{\pi} \theta \). These lines correspond to the rays defined by \( \theta = \frac{\pi}{4} c \), where \( c \) is a constant. Visually, these lines form a series of evenly spaced angles, denoting constant potential values. They are an essential part of understanding how potential is distributed in the complex plane, providing insight into regions of equal energy or field intensity. This perspective aids in comprehending overall patterns and behaviors within the domain in question.
Here, the equipotential lines are derived from the expression \( \phi(r, \theta) = \frac{4}{\pi} \theta \). These lines correspond to the rays defined by \( \theta = \frac{\pi}{4} c \), where \( c \) is a constant. Visually, these lines form a series of evenly spaced angles, denoting constant potential values. They are an essential part of understanding how potential is distributed in the complex plane, providing insight into regions of equal energy or field intensity. This perspective aids in comprehending overall patterns and behaviors within the domain in question.
Other exercises in this chapter
Problem 3
For \(w=z^{2}, u=x^{2}-y^{2}\) and \(v=2 x y .\) If \(x y=1, v=2\) and so the hyperbola \(x y=1\) is mapped onto the line \(v=2.\)
View solution Problem 4
\(f(z)=z+\operatorname{Ln} z+1\) is analytic for all \(z\) except \(z=y \leq 0 .\) But \(f^{\prime}(z)=1+1 / z\) and \(1+1 / z=0\) for \(z=-1\) Therefore \(f(z)
View solution Problem 5
\(S^{-1}(T(z))=\frac{a z+b}{c z+d}\) where $$\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]=\operatorname{adj}\left(\left[\begin{array}{rr} i & 1 \\
View solution Problem 5
\(f(z)=\left(z^{2}-1\right)^{1 / 2}=e^{\frac{1}{2} \operatorname{Ln}\left(z^{2}-1\right)}\) is analytic for all \(z\) outside the interval [-1,1] on the real ax
View solution