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Find the real and imaginary parts \(u\) and \(v\) of the given complex function \(f\) as functions of \(x\) and \(y\).\(f(z)=z+\frac{1}{z}\)

Short Answer

Expert verified
\(u(x, y) = x + \frac{x}{x^2 + y^2}\), \(v(x, y) = y - \frac{y}{x^2 + y^2}\).

Step by step solution

01

Set the Complex Number

Let the complex number be expressed as \(z = x + iy\), where \(x\) is the real part and \(iy\) is the imaginary part.
02

Calculate the Inverse

The inverse of \(z\) is \(\frac{1}{z}\). To find this, multiply \(z\) and its conjugate \(x - iy\). This gives: \[\frac{1}{z} = \frac{x - iy}{x^2 + y^2}.\]
03

Simplify the Expression

Substitute \(z = x + iy\) and \(\frac{1}{z} = \frac{x - iy}{x^2 + y^2}\) back into the function \(f(z)\): \[f(z) = z + \frac{1}{z} = (x + iy) + \left(\frac{x}{x^2+y^2} - \frac{iy}{x^2+y^2}\right).\]
04

Combine Real Parts

Combine the real parts of the expression: \[ u(x, y) = x + \frac{x}{x^2 + y^2}.\]
05

Combine Imaginary Parts

Combine the imaginary parts of the expression: \[ v(x, y) = y - \frac{y}{x^2 + y^2}.\]
06

Write Final Answer

The real part \(u\) and the imaginary part \(v\) of the complex function \(f(z)\) as functions of \(x\) and \(y\) are: \[ u(x, y) = x + \frac{x}{x^2 + y^2}, \quad v(x, y) = y - \frac{y}{x^2 + y^2}. \]

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Key Concepts

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

Complex Functions
Complex functions are a staple in the fascinating world of complex analysis. They map complex numbers to other complex numbers. A complex function might take a form like this: \(f(z) = z + \frac{1}{z}\). Here, \(z\) is a complex number often expressed as \(z = x + iy\), where \(x\) is the real part and \(iy\) is the imaginary part.

A unique aspect of complex functions is that they can be split into two parts:- A real part, which only uses real numbers- An imaginary part, which involves the imaginary unit \(i\).The complex function can therefore be denoted in terms of its real \(u\) and imaginary \(v\) parts: \[f(z) = u(x, y) + iv(x, y)\].

These functions have extensive applications in physics, engineering, and signal processing among other fields. Understanding complex functions begins with grasping how to separate them into their real and imaginary components.
Real and Imaginary Parts
Understanding the real and imaginary parts of a complex function is crucial as it allows us to analyze and manipulate these components separately. Let's focus on the function \(f(z) = z + \frac{1}{z}\). To dissect this function, we express \(z\) as \(x + iy\).

The next step involves calculating \(\frac{1}{z}\), the inverse of \(z\), using its conjugate. When you multiply \(z\) by its conjugate \(x - iy\), the imaginary parts cancel out, simplifying the denominator to \(x^2 + y^2\). Thus, the inverse is:\[\frac{1}{z} = \frac{x - iy}{x^2 + y^2}\]. Inserting \(z\) and \(\frac{1}{z}\) back into \(f(z)\), you have:\[f(z) = (x + iy) + \left(\frac{x}{x^2+y^2} - \frac{iy}{x^2+y^2}\right)\].Breaking this down further, the real part \(u(x, y)\) becomes:\[u(x, y) = x + \frac{x}{x^2 + y^2}\], and the imaginary part \(v(x, y)\) becomes:\[v(x, y) = y - \frac{y}{x^2 + y^2}\].

These separate parts help to clarify the overall behavior of the complex function in relation to its real and imaginary components.
Inverse of Complex Numbers
The inverse of a complex number \(z\) is a fundamental concept in complex analysis. To find the inverse \(\frac{1}{z}\), you multiply \(z\) by its conjugate. A complex number conjugate flips the sign of the imaginary component. With \(z = x + iy\), its conjugate is \(x - iy\).

When you multiply \(z\) by its conjugate, the imaginary parts \(iy\) and \(-iy\) cancel each other out. This results in:\[z \cdot \text{conjugate}(z) = (x + iy)(x - iy) = x^2 - (iy)^2\]. Since \(i^2 = -1\), this simplifies to:\[x^2 + y^2\].

Thus, for the inverse of \(z = x + iy\), expressed as \(\frac{1}{z}\), you use:\[\frac{x - iy}{x^2 + y^2}\].
The inverse operation is essential when functions involve division by a complex number, as it helps stabilize the expression by ensuring a real denominator. This fundamental concept aids in the simplification and deeper understanding of complex functions and their behavior.

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Most popular questions from this chapter

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.

Consider the multiple-valued function \(F(z)=(z-1+i)^{1 / 2}\). (a) What is the branch point of \(F ?\) Explain. (b) Explicitly define two distinct branches of \(f_{1}\) and \(f_{2}\) of \(F .\) In each case, state the branch cut.

Find the streamlines of the planar flow associated with the given complex function \(f\) and (b) sketch the streamlines.\(f(z)=i z\)

Consider the multiple-valued function \(F(z)=z^{1 / 3}\) that assigns to \(z\) the set of three cube roots of \(z\). Explicitly define three distinct branches \(f_{1}, f_{2}\), and \(f_{3}\) of \(F\), all of which have the nonnegative real axis as a branch cut.

Plot the images of the complex numbers \(z=1,1+i, 1-i\), and \(i\) under the given function \(f\) as position vectors, and (b) plot the images as vectors in the vector field associated with \(f\).\(f(z)=z-\frac{1}{z}\)

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