Problem 116 - Algebra and Trigonometry Real Mathematics, Real People

Question

Use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$\left[2\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)\right]^{12}$$

Step-by-Step Solution

Answer

The answer to the task \([2(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2})]^{12}\) in standard form is \(4096 + 0i\)

Step 1: Apply DeMoivre's Theorem

DeMoivre's Theorem states that for any complex number in form of \( r(\cos \theta+ i \sin \theta) \) and an integer \(n\), \([r(\cos \theta+ i \sin \theta)]^n = r^n(\cos n \theta+ i \sin n \theta)\). In our task, \(r=2\), \(\theta = \pi/2\), \(n=12\). So, the expression \([2(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2})]^{12}\) simplifies to \(2^{12} (\cos 12* \frac{\pi}{2} + i \sin 12* \frac{\pi}{2})\)

Step 2: Calculate powers

Calculate powers of 2 and simplify the expressions inside cos and sin. Hence, the expression will become \(4096 (\cos 6 \pi + i \sin 6 \pi)\).

Step 3: Calculate cosine and sine values

Since \(cos 2 \pi=1\), \(cos 6 \pi = 1\) and \(sin 2 \pi=0\), \(sin 6 \pi = 0\). So, this simplifies to \(4096(1+i0)\)

Step 4: Return to Standard Form

Last step it to return to standard form \(a + bi\). The standard form will be \(4096 + 4096*0i\) or \(4096 + 0i\)

Key Concepts

Complex Numbers

Complex numbers extend the idea of the one-dimensional number line by introducing imaginary numbers to form a two-dimensional plane, often referred to as the complex plane. In this plane, each complex number is expressed in the form of \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit defined by \(i^2 = -1\). Here:

• \(a\) is the real part of the complex number.
• \(b\) is the imaginary part.

Considering a complex number like \(2 + 3i\), it represents a point on the complex plane, where \(2\) is the horizontal component and \(3i\) is the vertical component. Understanding complex numbers is crucial, as they form the foundation for more advanced mathematical concepts, including DeMoivre's Theorem.

Standard Form

The standard form of a complex number is simply \(a + bi\), where \(a\) and \(b\) are real numbers. Converting complex numbers to this form makes it easier to perform arithmetic operations such as addition and subtraction.
In terms of DeMoivre's Theorem, once you calculate the power of a complex number in trigonometric form, you'll generally convert it back to the standard form for simplicity. For instance, if you have a complex number \(4096(1 + i0)\), the equivalent standard form is \(4096 + 0i\) or just \(4096\) since the imaginary part is zero. This format is practical for understanding the magnitude and direction of the number on the complex plane.

Trigonometric Form

The trigonometric form of a complex number allows for an easier way to raise complex numbers to a power, which is particularly useful in conjunction with DeMoivre’s Theorem.
The trigonometric form expresses a complex number as \(r(\cos \theta + i \sin \theta)\), where:

• \(r\) is the modulus or magnitude of the complex number.
• \(\theta\) is the argument or angle of the complex number.

The conversion from standard form to trigonometric form gives us \(r = \sqrt{a^2 + b^2}\) and \(\theta = \tan^{-1}(b/a)\). This form is powerful when performing operations involving powers and roots of complex numbers, as seen in the given exercise, where a complex number \([2(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2})]\) was raised to the twelfth power efficiently.

Powers of Complex Numbers

Using DeMoivre's Theorem is an efficient method for computing powers of complex numbers in trigonometric form. The theorem states that for any complex number \(r(\cos \theta + i \sin \theta)\) and an integer \(n\), \([r(\cos \theta + i \sin \theta)]^n = r^n(\cos n\theta + i \sin n\theta)\).
This provides a straightforward approach to finding the power of a complex number by multiplying the magnitude by itself \(n\) times and adjusting the argument of the angle accordingly by multiplying it by \(n\).
In the exercise, \([2(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})]^{12}\) is converted using DeMoivre's Theorem to \(4096(\cos 6\pi + i\sin 6\pi)\). Subsequently, since \(\cos 6\pi = 1\) and \(\sin 6\pi = 0\), it simplifies to \(4096(1 + 0i)\) or just \(4096\), demonstrating the ease of calculating high powers of complex numbers using this method.