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Define $$ e^{\theta i}=\cos \theta+i \sin \theta $$ Describe \(e^{\theta i}\) geometrically. Is \(\left|e^{\theta i}\right|=1 ?\)

Short Answer

Expert verified
Yes, \(|e^{\theta i}| = 1\) as it lies on the unit circle.

Step by step solution

01

Understanding the Formula

The given expression, \( e^{\theta i} = \cos \theta + i \sin \theta \), is Euler's formula. It's a way to represent complex numbers in exponential form.
02

Geometric Representation of \(e^{\theta i}\)

In the complex plane, \( e^{\theta i} \) can be seen as a point on the unit circle. The angle \( \theta \) is measured from the positive x-axis, corresponding to the real part (\(\cos \theta\)), with \(\sin \theta\) as the imaginary part.
03

Analyzing Magnitude

To find the magnitude of \( e^{\theta i} \), calculate the modulus: \(|e^{\theta i}| = \sqrt{(\cos \theta)^2 + (\sin \theta)^2} = \sqrt{1} = 1 \).
04

Conclusion

Since the magnitude of \( e^{\theta i} \) is 1, this confirms that \( e^{\theta i} \) lies on the unit circle in the complex plane. Its magnitude is always 1, regardless of \( \theta \).

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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 concept in mathematics, allowing for the extension of real numbers into a new dimension. A complex number is expressed in the form of \( a + bi \), where \( a \) is the real part and \( b \) is the imaginary part, with \( i \) being the imaginary unit. The imaginary unit \( i \) satisfies the equation \( i^2 = -1 \).
Understanding complex numbers enriches our comprehension of various mathematical phenomena, particularly those involving oscillations and waves. When dealing with complex numbers, Euler's formula becomes a crucial tool. It expresses a complex number in exponential form as \( e^{\theta i} = \cos \theta + i \sin \theta \). This representation simplifies multiplication and division operations in the complex plane, as it allows angles and magnitudes to be easily modified.
Unit Circle
The unit circle is an essential concept that connects trigonometry and complex numbers. It is defined as a circle with a radius of 1 centered at the origin of a coordinate system. In the context of complex numbers, any point \( e^{\theta i} \) on the unit circle can be represented as \( \cos \theta + i \sin \theta \).
This representation arises from Euler's formula and describes how one can plot complex exponentials.
  • The angle \( \theta \) is measured from the positive real axis, typically in radians.
  • The cosine of \( \theta \) gives the \( x \)-coordinate (real part).
  • The sine of \( \theta \) provides the \( y \)-coordinate (imaginary part).
Thus, every point on the unit circle can also be considered a complex number with unit magnitude, making it a perfect representation for periodic functions such as sine and cosine within the complex plane.
Magnitude in Complex Plane
Magnitude in the complex plane refers to the "length" of a complex number from the origin to its point \( (a, b) \) in the plane. For a complex number \( z = a + bi \), the magnitude is given by the formula \( |z| = \sqrt{a^2 + b^2} \). The magnitude signifies how far a particular point is from the origin in the complex plane.In our specific case, Euler's formula \( e^{\theta i} = \cos \theta + i \sin \theta \) gives us that the magnitude is always 1:
  • Apply the formula: \(|e^{\theta i}| = \sqrt{(\cos \theta)^2 + (\sin \theta)^2}\)
  • Effectively: \(|e^{\theta i}| = \sqrt{1}\)
  • This simplifies to \( |e^{\theta i}| = 1 \)
This consistent magnitude of 1 means that \( e^{\theta i} \) will always lie on the unit circle, demonstrating its unitary nature despite varying angles \( \theta \). This property makes it an incredibly powerful tool for understanding rotations and periodic motion in the complex plane.

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Most popular questions from this chapter

Prove (for \(E^{2}\) and \(E^{3}\) ) that $$ \bar{x} \cdot \bar{y}=|\bar{x}||\bar{y}| \cos \alpha $$ where \(\alpha\) is the angle between the vectors \(\overrightarrow{0 x}\) and \(\overrightarrow{0 y}\); we denote \(\alpha\) by \(\langle\bar{x}, \bar{y}\rangle .\)

Let $$ \begin{aligned} z &=r(\cos \theta+i \sin \theta) \\ z^{\prime} &=r^{\prime}\left(\cos \theta^{\prime}+i \sin \theta^{\prime}\right), \text { and } \\ z^{\prime \prime} &=r^{\prime \prime}\left(\cos \theta^{\prime \prime}+i \sin \theta^{\prime \prime}\right) \end{aligned} $$ as in Corollary 2. Prove that \(z=z^{\prime} z^{\prime \prime}\) if $$ r=|z|=r^{\prime} r^{\prime \prime}, \text { i.e., }\left|z^{\prime} z^{\prime \prime}\right|=\left|z^{\prime}\right|\left|z^{\prime \prime}\right|, \text { and } \theta=\theta^{\prime}+\theta^{\prime \prime} . $$ Discuss the following statement: To multiply \(z^{\prime}\) by \(z^{\prime \prime}\) means to rotate \(\overrightarrow{0 z^{\prime}}\) counterclockwise by the angle \(\theta^{\prime \prime}\) and to multiply it by the scalar \(r^{\prime \prime}=\) \(\left|z^{\prime \prime}\right| .\) Consider the cases \(z^{\prime \prime}=i\) and \(z^{\prime \prime}=-1 .\)

Prove the following about lines and line segments. (i) Show that any line segment in \(E^{n}\) is a bounded set, but the entire line is not. (ii) Prove that the diameter of \(L(\bar{a}, \bar{b})\) and of \((\bar{a}, \bar{b})\) equals \(\rho(\bar{a}, \bar{b})\).

Prove that $$ \overline{z+z^{\prime}}=\bar{z}+\overline{z^{\prime}} \text { and } \overline{z z^{\prime}}=\bar{z} \cdot \overline{z^{\prime}} . $$ Hence show by induction that $$ \overline{z^{n}}=(\bar{z})^{n}, n=1,2, \ldots, \text { and } \sum_{k=1}^{n} a_{k} z^{k}=\sum_{k=1}^{n} \bar{a}_{k} \bar{z}^{k} . $$

Let \(f: E^{1} \rightarrow E^{1}\) be given by $$ f(x)=\frac{1}{x} \text { if } x \neq 0, \text { and } f(0)=0. $$ Show that \(f\) is bounded on an interval \([a, b]\) iff \(0 \notin[a, b] .\) Is \(f\) bounded on (0,1)\(?\)

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