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Chapter 2: Problem 17

Find the real and imaginary parts \(u\) and \(v\) of the given complex function \(f\) as functions of \(r\) and \(\theta\).\(f(z)=\bar{z}\)

Short Answer

Expert verified
The real part is \(u = r\cos \theta\) and the imaginary part is \(v = -r\sin \theta\).

Step by step solution

01

Express complex number in polar form

The complex number \(z\) can be expressed in polar form as \(z = re^{i\theta}\). The conjugate of \(z\) is \(\bar{z} = re^{-i\theta}\).
02

Write the conjugate in rectangular form

The conjugate \(\bar{z} = re^{-i\theta}\) can be expanded using Euler's formula: \(\bar{z} = r(\cos(-\theta) + i\sin(-\theta)) = r(\cos \theta - i\sin \theta)\).
03

Identify real part \(u\) and imaginary part \(v\)

From \(\bar{z} = r(\cos \theta - i\sin \theta)\), we identify the real and imaginary parts as: \(u = r\cos \theta\) and \(v = -r\sin \theta\).

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Key Concepts

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

Polar Form
Complex numbers can be written in what we call polar form. This way of representing a complex number involves two parts: the magnitude and the angle.
Instead of using the standard form with real and imaginary components, denoted as \(z = a + bi\), we use the magnitude \(r\) and the angle \(\theta\).
The polar form is expressed as \(z = re^{i\theta}\).
  • \(r\) is the magnitude or modulus of the complex number.
  • \(\theta\) is the angle, usually in radians, from the positive real axis.
Why use polar form? It鈥檚 particularly useful in simplifying multiplication and division of complex numbers and in solving complex equations. In our problem, we used the polar form to express the conjugate of a complex number.
Euler's Formula
Euler's formula is a fundamental bridge between trigonometry and complex numbers. It states that for any real number \(\theta\), the complex exponential \(e^{i\theta}\) can be expressed as:\[e^{i\theta} = \cos \theta + i\sin \theta\]This formula helps relate complex exponentials to trigonometric functions, making it easier to manipulate expressions in polar form.
When you encounter a complex conjugate, as in the problem \(\bar{z} = re^{-i\theta}\), applying Euler鈥檚 formula transforms it into a mixture of cosine and sine:
  • \(\cos(-\theta) = \cos \theta\)
  • \(\sin(-\theta) = -\sin \theta\)
This simplification is critical for identifying the real and imaginary components, transforming the expression into something far easier to work with.
Real and Imaginary Parts
The real and imaginary parts are crucial to understanding how complex numbers work in different forms. In standard form, a complex number \(z = a + bi\) is split into two parts: the real part \(a\) and the imaginary part \(bi\).
In our problem, we determined these parts by expanding the polar form into a trigonometric expression:
  • The real part \(u = r\cos \theta\)
  • The imaginary part \(v = -r\sin \theta\)
These correspond respectively to the cosine and sine components derived from Euler's formula for the conjugate of \(z\). Understanding these parts allows us to handle complex number operations, like addition and multiplication, when they are expressed in their rectangular form. This insight is fundamental in a wide range of applications, from electrical engineering to wave physics.

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Most popular questions from this chapter

Use a CAS to show that the given function is not continuous inside the unit circle by plotting the image of the given continuous parametric curve. (Be careful, Mathematica and Maple plots can sometimes be misleading.)\(f(z)=z+\operatorname{Arg}(z), z(t)=-\frac{1}{2}+\frac{1}{2} \sqrt{3} i t,-1 \leq t \leq 1\)

Groups of Isometries In this project we investigate the relationship between complex analysis and the Euclidean geometry of the Cartesian plane. The Euclidean distance between two points \(\left(x_{1}, y_{1}\right)\) and \(\left(x_{2}, y_{2}\right)\) in the Cartiesian plane is $$ d\left(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right)\right)=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} $$ Of course, if we consider the complex representations \(z_{1}=x_{1}+i y_{1}\) and \(z_{2}=\) \(x_{2}+i y_{2}\) of these points, then the Euclidean distance is given by the modulus $$ d\left(z_{1}, z_{2}\right)=\left|z_{2}-z_{1}\right| $$ A function from the plane to the plane that preserves the Euclidean distance between every pair of points is called a Euclidean isometry of the plane. In particular, a complex mapping \(w=f(z)\) is a Euclidean isometry of the plane if $$ \left|z_{2}-z_{1}\right|=\left|f\left(z_{1}\right)-f\left(z_{2}\right)\right| $$ for every pair of complex numbers \(z_{1}\) and \(z_{2}\).(a) Prove that every linear mapping of the form \(f(z)=a z+b\) where \(|a|=1\) is a Euclidean isometry. A group is an algebraic structure that occurs in many areas of mathematics. A group is a set \(G\) together with a special type of function \(*\) from \(G \times G\) to \(G\). The function \(*\) is called a binary operation on \(G\), and it is customary to use the notation \(a * b\) instead of \(*(a, b)\) to represent a value of \(* .\) We now give the formal definition of a group. A group is a set \(G\) together with a binary operation \(*\) on \(G\), which satisfies the following three properties: (i) for all elements \(a, b\), and \(c\) in \(G, a *(b * c)=(a * b) * c\),(b) Prove that composition of functions is a binary operation on Isom \(_{+}(\mathbf{E})\). That is, prove that if \(f\) and \(g\) are functions in Isom \(_{+}(\mathbf{E})\), then the function \(f \circ g\) defined by \(f \circ g(z)=f(g(z))\) is an element in Isom \(_{+}(\mathbf{E})\). (c) Prove that the set Isom \(_{+}(\mathbf{E})\) with composition satisfies property \((i)\) of a group. (d) Prove that the set \(\operatorname{Isom}_{+}(\mathbf{E})\) with composition satisfies property \((i i)\) of a group. That is, show that there exists a function \(e\) in Isom \(_{+}(\mathbf{E})\) such that \(e \circ f=f \circ e=f\) for all functions \(f\) in Isom \(_{+}(\mathbf{E})\). (e) Prove that the set Isom \(_{+}(\mathbf{E})\) with composition satisfies property \((i i i)\) of a group.

Find the image of the given set under the principal square root mapping \(w=z^{1 / 2}\). Represent the mapping by drawing the set and its image.the parabola \(x=\frac{9}{4}-\frac{y^{2}}{9}\)

Show that the image of the line \(x=k, x \neq 0\), under the reciprocal map defined on the extended complex plane is the circle \(\left|w-\frac{1}{2 k}\right|=\left|\frac{1}{2 k}\right|\).

Find three sets in the complex plane that map onto the set \(\arg (w)=\pi\) under the mapping \(w=z^{3}\).
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