Chapter 9: Problem 1
Express \((\cos \theta+j \sin \theta)^{9}\) and \((\cos \theta+j \sin \theta)^{1 / 2}\) in the form \(\cos n \theta+\mathrm{j} \sin n \theta\).
Short Answer
Expert verified
Expressed as \(\cos(9\theta) + j \sin(9\theta)\) and \(\cos\left(\frac{1}{2}\theta\right) + j \sin\left(\frac{1}{2}\theta\right).\)
Step by step solution
01
Understand De Moivre's Theorem
De Moivre's Theorem states that for any complex number in the form \(\cos \theta + j \sin \theta\), raising it to a power \(n\) results in another complex number \(\cos(n\theta) + j \sin(n\theta)\). So, \((\cos \theta + j \sin \theta)^n = \cos(n\theta) + j \sin(n\theta)\).
02
Apply the Theorem for the Exponent 9
Given \((\cos \theta + j \sin \theta)^9\), according to De Moivre's Theorem, \(n = 9\). Therefore, \((\cos \theta + j \sin \theta)^9 = \cos(9\theta) + j \sin(9\theta)\).
03
Apply the Theorem for the Square Root
For \((\cos \theta + j \sin \theta)^{1/2}\), let \(n = \frac{1}{2}\). Again, using De Moivre's Theorem, we get \((\cos \theta + j \sin \theta)^{1/2} = \cos\left(\frac{1}{2}\theta\right) + j \sin\left(\frac{1}{2}\theta\right)\).
04
Conclude with General Forms
Now that both expressions have been rewritten using De Moivre's Theorem: \((\cos \theta + j \sin \theta)^9\) becomes \(\cos(9\theta) + j \sin(9\theta)\) and \((\cos \theta + j \sin \theta)^{1/2}\) becomes \(\cos\left(\frac{1}{2}\theta\right) + j \sin\left(\frac{1}{2}\theta\right).\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Complex Numbers
Complex numbers are numbers that have both a real and imaginary part, usually represented as \(a + bj\), where \(a\) is the real part and \(bj\) is the imaginary part. Here, \(j\) is the imaginary unit where \(j^2 = -1\). This concept allows for an extended understanding and handling of numbers that go beyond the traditional number line, encapsulating more dimensions of algebra.
- Real Part: In the expression \(a + bj\), the "\(a\)" component represents the real part, much like any ordinary number you might find on the number line.
- Imaginary Part: The "\(bj\)" portion is termed imaginary because it contains the imaginary unit \(j\), which taps into a dimension not captured by real numbers alone.
The Trigonometric Form of Complex Numbers
When dealing with complex numbers, the trigonometric form offers a unique and beneficial way to visualize and manipulate these numbers. This form is particularly useful for operations like raising numbers to powers or finding roots. In trigonometric form, a complex number is expressed as \(r(\cos \theta + j \sin \theta)\), where \(r\) is the modulus and \(\theta\) is the argument of the complex number.
- Modulus (\(r\)): The modulus is the distance from the origin to the point representing the complex number in the complex plane. It can be calculated using the formula \(r = \sqrt{a^2 + b^2}\).
- Argument (\(\theta\)): The argument is the angle formed with the positive real axis and this can be found using \(\tan^{-1}(b/a)\).
Calculating Powers and Roots Using De Moivre's Theorem
De Moivre's Theorem is a powerful tool that simplifies calculations of powers and roots of complex numbers in trigonometric form. The theorem asserts that given a complex number in the form \(r(\cos \theta + j \sin \theta)\), raising it to a power \(n\) is as simple as computing \(r^n(\cos(n\theta) + j \sin(n\theta))\), thereby transforming operations into more manageable forms.For example:
- Exponentiation: To raise a complex number like \((\cos \theta + j \sin \theta)^9\), simply multiply the angle \(\theta\) by 9, resulting in \(\cos(9\theta) + j \sin(9\theta)\).
- Extracting Roots: Finding roots involves taking the root of the modulus, while dividing the angle. For instance, \((\cos \theta + j \sin \theta)^{1/2}\) becomes \(\cos(\frac{\theta}{2}) + j \sin(\frac{\theta}{2})\).