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Show that when \(z\) is represented by polar coordinates, the derivative of a function \(f(z)\) can be written as $$\frac{d f}{d z}=e^{-i \theta}\left(\frac{\partial U}{\partial r}+i \frac{\partial V}{\partial r}\right),$$ where \(U\) and \(V\) are the real and imaginary parts of \(f(z)\) written in polar coordinates. What are the C-R conditions in polar coordinates? Hint: Start with the C-R conditions in Cartesian coordinates and apply the chain rule to them using \(r=\sqrt{x^{2}+y^{2}}\) and \(\theta=\tan ^{-1}(y / x)=\cos ^{-1}\left(x / \sqrt{x^{2}+y^{2}}\right)\).

Short Answer

Expert verified
The derivative of a function in polar coordinates is given by \(\frac{d f}{d z}=e^{-i \theta}\left(\frac{\partial U}{\partial r}+i\frac{\partial V}{\partial r}\right)\). The Cauchy-Riemann conditions in polar coordinates result in: \(r\frac{\partial U}{\partial r} = \frac{\partial V}{\partial \theta}\) and \(\frac{\partial U}{\partial \theta} + r\frac{\partial V}{\partial r} = 0\).

Step by step solution

01

Understand the Variables and Functions

To begin with, focus on understanding the different parts of the formula. The function \(f(z)\) is a complex-valued function of a complex variable, written in the form \(f(z)=U(r,\theta)+iV(r,\theta)\) where \(U(r,\theta)\) and \(V(r,\theta)\) are real-valued functions. The variable \(z\) is also represented in polar coordinates as \(z=re^{i\theta}\), where \(r\) is the magnitude and \(\theta\) is the phase angle.
02

Write Down the Cauchy-Riemann equations in Cartesian Coordinates

In Cartesian coordinates, the Cauchy-Riemann (C-R) conditions state that if \(f(z)=u(x,y)+iv(x,y)\), then \(u\) and \(v\) must satisfy: \(\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\) and \(\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}\)
03

Convert Cartesian coordinates to Polar

Transforming from Cartesian to polar coordinates, the expressions become \(x=r\cos(\theta)\), \(y=r\sin(\theta)\), \(\frac{\partial}{\partial x}=\cos(\theta)\frac{\partial}{\partial r}-\frac{\sin(\theta)}{r}\frac{\partial}{\partial \theta}\), and \(\frac{\partial}{\partial y}=\sin(\theta)\frac{\partial}{\partial r}+\frac{\cos(\theta)}{r}\frac{\partial}{\partial \theta}\)
04

Apply substitutions to the C-R conditions

Apply the conversion to the C-R conditions from Step 2 using the results from Step 3. Combine like terms and simplify.
05

Derive the derivative in Polar Coordinates

Next, derive the formula for the derivative of a function in polar coordinates using the C-R conditions derived in Step 4. Apply the euler's formula: \(z=re^{i\theta}\) for the derivative. After some calculations, you will arrive at the formula given in the exercise.
06

Verify the C-R conditions in Polar Coordinates

Finally, check the C-R conditions in polar coordinates derived from Step 4. The resulting formulae should be the real and imaginary parts of the derivative formula obtained in Step 5.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cauchy-Riemann conditions
The Cauchy-Riemann (C-R) conditions are a set of equations that are essential in complex analysis. They act as a criterion for differentiability of a complex function. Specifically, for a function to be differentiable at a point, it must satisfy these conditions at that point. For a complex function written as f(z) = U(x, y) + iV(x, y) where U and V are the real and imaginary parts, respectively, the C-R conditions in Cartesian coordinates are:

  • \(\frac{\partial U}{\partial x} = \frac{\partial V}{\partial y}\)
  • \(\frac{\partial U}{\partial y} = -\frac{\partial V}{\partial x}\)

These equations imply that the partial derivatives of the real and imaginary parts of the function must satisfy certain relationships. These conditions not only imply that a function is differentiable but also that derivatives are conformal, meaning they preserve angles and shapes locally. The relationship between the partial derivatives in the C-R conditions illustrates the interconnectedness of the real and imaginary components of complex functions.
Polar coordinate transformation
Transforming Cartesian coordinates into polar coordinates is a key concept in complex analysis and physics due to its utility in simplifying certain types of problems, especially those involving circular or rotational symmetry. In polar coordinates, a point in the plane is described by the distance from the origin r and the angle \(\theta\) formed with the positive x-axis.

The transformation equations are:
  • \(x = r\cos(\theta)\)
  • \(y = r\sin(\theta)\)

Polar coordinates offer a natural way to describe complex numbers, where any complex number 'z' can be written as z = \(re^{i\theta}\). In terms of calculus, differentiating and integrating functions becomes intriguing as one must account for the radial and angular components. Thus, the application of polar coordinates invokes the need for different forms of the C-R conditions, chain rule, and understanding partial derivatives in a transformed coordinate system.
Complex analysis
Complex analysis is a branch of mathematics that focuses on functions of complex variables. It brings together many concepts of calculus and algebra, featuring a rich structure and profound results that have implications across other fields such as engineering, physics, and number theory. One of the key tools in complex analysis is the study of complex derivatives, which leads to deep results like the Cauchy integral formula and Liouville's theorem.

Within complex analysis, differentiability is a stringent condition, linked to the C-R conditions that help in identifying functions that are not just differentiable but also holomorphic (complex-analytic). These functions have unique properties, such as having power series representations around points where they are differentiable. Understanding how to manipulate complex functions, including finding derivatives in polar coordinates as outlined in the exercise, is a fundamental aspect of complex analysis.
Chain rule in calculus
The chain rule is a formula to compute the derivative of a composite function. In basic calculus, if a function y depends on u which itself depends on x, then the derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x. Symbolically, it can be written as:
\(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)
In complex analysis, the chain rule becomes a powerful tool when we transform between different coordinate systems, such as from Cartesian to polar coordinates. It allows us to relate partial derivatives in one system to those in another. This is vital when applying the C-R conditions to functions expressed in polar coordinates, as seen in the exercise. In polar coordinates, this takes a little nuance as we deal with radial and angular components, emphasizing the beauty and complexity of calculus in different dimensions.

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Most popular questions from this chapter

Use the definition of an integral as the limit of a sum and the fact that absolute value of a sum is less than or equal to the sum of absolute values to prove the Darboux inequality.

Let \(f(t)=u(t)+i v(t)\) be a (piecewise) continuous complex-valued function of a real variable \(t\) defined in the interval \(a \leq t \leq b\). Show that if \(F(t)=U(t)+i V(t)\) is a function such that \(d F / d t=f(t)\), then $$\int_{a}^{b} f(t) d t=F(b)-F(a) .$$ This is the fundamental theorem of calculus for complex variables.

Let \(f\) be analytic within and on the circle \(\gamma_{0}\) given by \(\left|z-z_{0}\right|=r_{0}\) and integrated in the positive sense. Show that Cauchy's inequality holds: $$\left|f^{(n)}\left(z_{0}\right)\right| \leq \frac{n ! M}{r_{0}^{n}}$$ where \(M\) is the maximum value of \(|f(z)|\) on \(\gamma_{0}\).

Obtain the first few nonzero terms of the Laurent series expansion of each of the following functions about the origin. Also find the integral of the function along a small simple closed contour encircling the origin. (a) \(\frac{1}{\sin z}\); (b) \(\frac{1}{1-\cos z}\); (c) \(\frac{z}{1-\cosh z}\); (d) \(\frac{z^{2}}{z-\sin z}\); (e) \(\frac{z^{4}}{6 z+z^{3}-6 \sinh z}\); (f) \(\frac{1}{z^{2} \sin z}\); (g) \(\frac{1}{e^{z}-1}\).

Prove the following identities. (a) \(\cos ^{-1} z=-i \ln \left(z \pm \sqrt{z^{2}-1}\right)\), (b) \(\sin ^{-1} z=-i \ln \left[i z \pm \sqrt{1-z^{2}}\right]\), (c) \(\tan ^{-1} z=\frac{1}{2 i} \ln \left(\frac{i-z}{i+z}\right)\), (d) \(\quad \cosh ^{-1} z=\ln \left(z \pm \sqrt{z^{2}-1}\right)\), (e) \(\sinh ^{-1} z=\ln \left(z \pm \sqrt{z^{2}+1}\right)\), (f) \(\tanh ^{-1} z=\frac{1}{2} \ln \left(\frac{1+z}{1-z}\right)\).

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