Chapter 3: Problem 55
Show that the line \(x=1\) on the \(z\) plane is transformed into the circle $$ u^{2}+v^{2}-u=0 $$ on the \(w\) plane by the function $$ w=(z-1) /(z+1) $$
Short Answer
Expert verified
The line \(x=1\) on the \(z\)-plane becomes the circle \(u^2 + v^2 - u = 0\) on the \(w\)-plane.
Step by step solution
01
Parametrize the Line
The line given in the problem is \(x = 1\) on the \(z\)-plane. We can write the complex number \(z\) in terms of its real and imaginary parts, \(z = x + yi\). For the line \(x=1\), we have \(z = 1 + yi\), where \(y\) is a real number.
02
Apply the Transformation Function
Substitute \(z = 1 + yi\) into the transformation function \(w = \frac{z-1}{z+1}\):\[w = \frac{(1 + yi) - 1}{(1 + yi) + 1} = \frac{yi}{2 + yi}.\]
03
Multiply by the Conjugate
To simplify \(w\), we multiply the numerator and the denominator by the conjugate of the denominator:\[ w = \frac{yi(2 - yi)}{(2 + yi)(2 - yi)} = \frac{2yi + y^2}{4 + y^2}.\]
04
Separate into Real and Imaginary Parts
Express \(w\) in the form \(w = u + vi\):\[ u = \frac{y^2}{4 + y^2}, \quad v = \frac{2y}{4 + y^2}.\]
05
Derive the Equation of the Circle
We want to show that these \(u\) and \(v\) satisfy the circle equation \(u^2 + v^2 - u = 0\):\[u^2 + v^2 = \left( \frac{y^2}{4 + y^2} \right)^2 + \left( \frac{2y}{4 + y^2} \right)^2 = \frac{y^4 + 4y^2}{(4 + y^2)^2},\]\[u = \frac{y^2}{4 + y^2}.\] Now, substitute into the circle's equation:\[ u^2 + v^2 - u = \frac{y^4 + 4y^2}{(4 + y^2)^2} - \frac{y^2}{4 + y^2} = \frac{y^4 + 4y^2 - y^2 (4 + y^2)}{(4 + y^2)^2} = 0.\]The numerator simplifies to zero, showing the identity holds.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Numbers
Complex numbers are a fascinating branch of mathematics that extends the concept of ordinary numbers. They consist of two parts: a real part and an imaginary part. The imaginary part is marked with an 'i', which is the square root of -1. Such numbers are written in the form: \[ z = a + bi \] where \(a\) is the real part, and \(bi\) is the imaginary part.
- Imaginary Unit: Represented by 'i', defined as \(i^2 = -1\).
- Real Part: Denoted as \(a\) in the complex number \(a + bi\).
- Imaginary Part: Denoted as \(b\) in the complex number \(a + bi\).
Parametrization
Parametrization is a powerful tool for describing geometric objects like lines or surfaces using variables, often making them easier to work with analytically. In our context, parametrization involves expressing the points on a line using a single real parameter. For example, the line \(x=1\) is straightforward in its representation. However, when described on the complex plane where \(z = x + yi\), it's necessary to express \(z\) using a parameter.
- Parameter: A variable, say \(y\), representing a continuum of values.
- Line In Complex Plane: Represents \(z\) as \(1 + yi\).
Conjugate Multiplication
Conjugate multiplication is a technique that helps in simplifying complex fractions, crucial for complex transformations. In complex numbers, the conjugate of \(z = a + bi\) is \(z^* = a - bi\). Multiplying the numerator and denominator of a complex fraction by the conjugate of the denominator simplifies the expression. Here's how it helps:
- Conjugate: For \(z = a + bi\), it's \(a - bi\).
- Rationalization: Multiplies by the conjugate to eliminate imaginary parts from the denominator.
Circle Equation
Circle equations describe the set of all points that are equidistant from a fixed center point in the complex plane or coordinate system. The standard form of a circle's equation is: \[ (x - h)^2 + (y - k)^2 = r^2 \] where \((h, k)\) is the center, and \(r\) is the radius. However, in transformations, the form might be different, as depicted by the equation of the circle in the exercise: \[ u^2 + v^2 - u = 0 \]
- Variants: Equations can vary in form due to transformations and different parameterizations.