Problem 10
Question
Exer. 1-12: Use De Moivre's theorem to change the given complex number to the form \(a+b i\), where \(a\) and \(b\) are real numbers. $$ \left(-\frac{\sqrt{3}}{2}-\frac{1}{2} i\right)^{50} $$
Step-by-Step Solution
Verified Answer
The complex number is \(\frac{1}{2} + i\frac{\sqrt{3}}{2}\).
1Step 1: Identify the Complex Number in Polar Form
Given the complex number \(-\frac{\sqrt{3}}{2}-\frac{1}{2} i\). To apply De Moivre's theorem, express it in polar form as \(r(\cos \theta + i \sin \theta)\). The given complex number has a magnitude \(r = 1\), since \((-\frac{\sqrt{3}}{2})^2 + (-\frac{1}{2})^2 = 1\). The argument \(\theta\) is \(-\frac{5\pi}{6}\), as determined by evaluating the angle corresponding to the cosine of \(-\frac{\sqrt{3}}{2}\) and sine of \(-\frac{1}{2}\).
2Step 2: Apply De Moivre's Theorem
De Moivre's theorem states: \[(r(\cos \theta + i \sin \theta))^n = r^n(\cos(n \theta) + i \sin(n \theta)).\]Apply it with \(r = 1\), \(\theta = -\frac{5\pi}{6}\), and \(n = 50\). So,\[1^{50} \left(\cos\left(50 \times -\frac{5\pi}{6}\right) + i \sin\left(50 \times -\frac{5\pi}{6}\right)\right).\]
3Step 3: Simplify the Arguments
Calculate \(50 \times -\frac{5\pi}{6} = -\frac{250\pi}{6} = -\frac{125\pi}{3}\). To find the equivalent angle within \([0, 2\pi)\), simplify this:Divide \(-\frac{125\pi}{3}\) by \(2\pi\) to find how it wraps around the circle:\[-\frac{125\pi}{3} \div 2\pi = -\frac{125}{6} = -20.8333\ldots\]Find the closest integer (-21), so the angle is:\[-\frac{125\pi}{3} + 21 \times 2\pi = \frac{\pi}{3}.\]
4Step 4: Calculate the Resulting Complex Number
Compute \(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\):\[\cos \frac{\pi}{3} = \frac{1}{2}, \quad \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}.\]Thus, the expression is:\[\frac{1}{2} + i\frac{\sqrt{3}}{2}.\]
5Step 5: Final Step: Express in Form \(a + bi\)
Combine the real and imaginary parts obtained:\[a + bi = \frac{1}{2} + i\frac{\sqrt{3}}{2}.\]This is the solution in the form of \(a + bi\).
Key Concepts
Complex NumbersPolar FormMagnitude and ArgumentTrigonometric Form
Complex Numbers
Complex numbers are an extension of real numbers. They are expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. The imaginary unit \(i\) is defined by the property \(i^2 = -1\).
Complex numbers allow for the extension of algebraic concepts beyond real numbers, especially useful in solving equations that have no real solutions. A complex number \(a + bi\) has:
Complex numbers allow for the extension of algebraic concepts beyond real numbers, especially useful in solving equations that have no real solutions. A complex number \(a + bi\) has:
- a real part, \(a\), and
- an imaginary part, \(bi\).
Polar Form
A complex number can also be expressed in polar form, which can sometimes simplify multiplication and division. The polar form is written as \(r(\cos \theta + i \sin \theta)\), where \(r\) is the magnitude and \(\theta\) is the argument.
Converting from the usual \(a + bi\) form to polar form involves:
Converting from the usual \(a + bi\) form to polar form involves:
- Calculating \(r = \sqrt{a^2 + b^2}\), the distance from the origin to the point in the complex plane.
- Finding \(\theta = \tan^{-1}(\frac{b}{a})\), the angle between the positive real axis and the line representing the complex number.
Magnitude and Argument
The magnitude, \(r\), and argument, \(\theta\), are critical components of a complex number in polar form. The magnitude represents the length of the vector from the origin to the point \((a, b)\) in the complex plane.
It is calculated using the formula: \(r = \sqrt{a^2 + b^2}\). The argument, on the other hand, is the angle \(\theta\) with the positive x-axis and is found by \(\theta = \tan^{-1}(\frac{b}{a})\).
In some cases, the argument needs to be adjusted depending on the quadrant where the complex number is located. Notably, when handling powers of complex numbers as in De Moivre's theorem, simplifying \(\theta\) to be within range \([0, 2\pi)\) is essential for clarity and simplicity.
It is calculated using the formula: \(r = \sqrt{a^2 + b^2}\). The argument, on the other hand, is the angle \(\theta\) with the positive x-axis and is found by \(\theta = \tan^{-1}(\frac{b}{a})\).
In some cases, the argument needs to be adjusted depending on the quadrant where the complex number is located. Notably, when handling powers of complex numbers as in De Moivre's theorem, simplifying \(\theta\) to be within range \([0, 2\pi)\) is essential for clarity and simplicity.
Trigonometric Form
Trigonometric form of complex numbers is synonymous with their polar form. It expresses complex numbers as \(r(\cos \theta + i \sin \theta)\), which closely relates to their geometric representation. This form is convenient for performing operations like multiplication and division, and it simplifies raising numbers to powers and finding roots.
The trigonometric form leverages the properties of trigonometric functions \(\cos\) and \(\sin\), making it easier to work with complex numbers by utilizing their periodic nature. Using De Moivre's theorem, calculations involving powers become straightforward: \[(r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i \sin(n\theta)).\]
Therefore, mastering this form is a powerful tool in complex number analysis, especially helpful in fields requiring extensive calculation, such as electrical engineering.
The trigonometric form leverages the properties of trigonometric functions \(\cos\) and \(\sin\), making it easier to work with complex numbers by utilizing their periodic nature. Using De Moivre's theorem, calculations involving powers become straightforward: \[(r(\cos \theta + i \sin \theta))^n = r^n(\cos(n\theta) + i \sin(n\theta)).\]
Therefore, mastering this form is a powerful tool in complex number analysis, especially helpful in fields requiring extensive calculation, such as electrical engineering.
Other exercises in this chapter
Problem 9
Solve \(\triangle A B C\). $$\gamma=115^{\circ} 10^{\prime}, \quad a=1.10, \quad b=2.10$$
View solution Problem 9
Solve \(\triangle A B C\). $$\gamma=53^{\circ} 20^{\prime}, \quad a=140, \quad c=115$$
View solution Problem 10
Exer. 9 -12: Show that the vectors are orthogonal. \(\langle 3,6\rangle\) \(\langle 4,-2\rangle\)
View solution Problem 10
Solve \(\triangle A B C\). $$\alpha=27^{\circ} 30^{\prime}, \quad c=52.8, \quad a=28.1$$
View solution