Chapter 14: Problem 5
Find end sketch the path and its orientation given by: $$z(t)=e^{i t}(0 \leq t \leq \pi)$$
Short Answer
Expert verified
The path is the upper semicircle from \( (1,0) \) to \( (-1,0) \), oriented counterclockwise.
Step by step solution
01
Understand the given complex function
The function given is \( z(t) = e^{it} \). This represents a point on the unit circle in the complex plane, where \( e^{it} \) is equivalent to \( \cos(t) + i\sin(t) \). For \( 0 \leq t \leq \pi \), this means we are considering the semicircle from \( t = 0 \) to \( t = \pi \).
02
Identify the path of the function
For \( z(t) = e^{i t} = \cos(t) + i \sin(t) \), as \( t \) increases from \( 0 \) to \( \pi \), the path traced is the upper half of the unit circle, starting from \( (1,0) \) (corresponding to \( t = 0 \)) and ending at \( (-1,0) \) (corresponding to \( t = \pi \)).
03
Determine the orientation of the path
The function \( z(t) = e^{i t} \) describes a movement counterclockwise along the unit circle as \( t \) increases. Therefore, the orientation of this path is counterclockwise from the point \( (1,0) \) to \( (-1,0) \).
04
Sketch the path
Draw the unit circle on the complex plane, highlighting the top half from the point \( (1, 0) \) to \( (-1, 0) \). Indicate the start point at \( (1,0) \), the end point at \( (-1,0) \), and show an arrow pointing counterclockwise to denote the orientation of the path.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Complex Plane
In complex analysis, the **complex plane** is an essential concept that provides a visual way to represent complex numbers. It is similar to the Cartesian plane, but instead of plotting ordered pairs of real numbers, each point on the plane represents a complex number.
The power of the complex plane lies in its ability to easily visualize complex functions, like in the given exercise. When we plot functions such as \( z(t) = e^{it} \) which is equivalent to \( \cos(t) + i\sin(t) \), we're visualizing the position on a circle defined by the sine (vertical) and cosine (horizontal) components as time \( t \)increases.
- The horizontal axis, known as the real axis, represents the real part of a complex number.
- The vertical axis, referred to as the imaginary axis, represents the imaginary part of a complex number.
The power of the complex plane lies in its ability to easily visualize complex functions, like in the given exercise. When we plot functions such as \( z(t) = e^{it} \) which is equivalent to \( \cos(t) + i\sin(t) \), we're visualizing the position on a circle defined by the sine (vertical) and cosine (horizontal) components as time \( t \)increases.
Unit Circle
The **unit circle** is a fundamental concept when examining functions like \( z(t) = e^{it} \) in the complex plane. Defined as a circle with a radius of 1 centered at the origin \((0,0)\), it holds special significance in both trigonometry and complex analysis.
- The unit circle allows us to represent complex exponentials (\( e^{it} \)) conveniently.
- For example, as \( t \) varies from 0 to \( \pi \), we traverse the top half of the circle, starting at \( (1,0) \) and ending at \( (-1,0) \).
Counterclockwise Orientation
In complex analysis, the **orientation** of a path is an important concept, especially when dealing with functions on the unit circle. For the function \( z(t) = e^{it} \), the path is traced in a counterclockwise direction as \( t \) increases. Understanding this orientation is crucial in visualizing and analyzing functions in the complex plane.
- A counterclockwise orientation means that, as \( t \) increases, the angle from the positive real axis moves leftwards around the unit circle.
- Starting from \( t = 0 \) at point \( (1,0) \), the path moves towards \( t = \pi \) at point \( (-1,0) \).
- This specific orientation is a standard practice in mathematics to denote the positive direction in circular paths, making it easier to define and interpret integrals involving complex functions.