Chapter 20: Problem 4
The analytic function \(f(z)=\cosh z\) is conformal except at \(z=\)
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Chapter 20: Problem 4
The analytic function \(f(z)=\cosh z\) is conformal except at \(z=\)
These are the key concepts you need to understand to accurately answer the question.
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Construct a linear fractional transformation that maps the given triple \(z_{1}, z_{2}\), and \(z_{3}\) to the triple \(w_{1}, w_{2}\), and \(w_{3}\). \(i, 0,-i\) to \(0,1, \infty\)
A curve in the \(z\) -plane and a complex mapping \(w=f(z)\) are given. In each case, find the image curve in the \(w\) -plane. Circular arc \(r=2,0 \leq \theta \leq \pi / 2\), under \(w=z^{1 / 2}\)
Verify that div \(\mathbf{F}=0\) and curl \(\mathbf{F}=\mathbf{0}\) for the given vector field \(\mathbf{F}(x, y)\) by examining the corresponding complex function \(g(z)=P(x, y)-i Q(x, y)\). Find a complex potential for the vector field and sketch the equipotential lines. \(\mathbf{F}(x, y)=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}} \mathbf{i}+\frac{2 x y}{\left(x^{2}+y^{2}\right)^{2}} \mathbf{j}\)
For any two real numbers \(k\) and \(x_{1}\), the function \(G(z)=\) \(k \operatorname{Ln}\left(z-x_{1}\right)\) is analytic in the upper half-plane and therefore is a complex potential for a flow. The real number \(x_{1}\) is called a sink when \(k<0\) and a source for the flow when \(k>0\). (a) Show that the streamlines are rays emanating from \(x_{1}\). (b) Show that \(\mathbf{V}=\left(k /\left|z-x_{1}\right|^{2}\right)\left(z-x_{1}\right)\) and conclude that the flow is directed toward \(x_{1}\) precisely when \(k<0\).
A frame for a membrane is defined by \(u\left(e^{i \theta}\right)=e^{-1 \theta \mid}\) for \(-\pi \leq \theta \leq \pi .\) Use the Poisson integral formula for the unit disk to estimate the equilibrium displacement of the membrane at \((-0.5,0),(0,0)\), and \((0.5,0)\).
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