Question

Compute \([\cos (3 \pi 2)+i \sin (3 \pi / 2)]^{6}\)

Short answer

The short answer for the given complex expression \([\cos (3 \pi / 2) + i \sin (3 \pi / 2)]^{6}\) is \(1\).

Step-by-step solution

Identify the relevant variables in the expression

The given expression is \([\cos (3 \pi / 2) + i \sin (3 \pi / 2)]^{6}\). In this case, we have \(\theta = 3 \pi / 2\) and we want to calculate the sixth power of this complex number, so n = 6.

Apply De Moivre's theorem

Using De Moivre's theorem, we need to calculate the expression \((\cos \theta + i \sin \theta)^n\), which will yield \(\cos n\theta + i \sin n\theta\) as the result. In this case, \(\theta = 3 \pi / 2\) and n = 6, so our expression becomes: \((\cos (3 \pi / 2) + i \sin (3 \pi / 2))^6 = \cos (6 \cdot 3 \pi / 2) + i \sin (6 \cdot 3 \pi / 2)\)

Calculate the values of the trigonometric functions

Now, we need to find the values of \(\cos (9 \pi)\) and \(\sin (9 \pi)\): \(\cos (9 \pi) = \cos (8 \pi + \pi) = 1\) \(\sin (9 \pi) = \sin (8 \pi + \pi) = 0\)

Substitute the values back into the expression

Finally, substitute the calculated values of the trigonometric functions back into the expression: \(\cos (9 \pi) + i \sin (9 \pi) = 1 + 0i\)

Write the final answer

The final answer for the given complex expression is: \([\cos (3 \pi / 2) + i \sin (3 \pi / 2)]^{6} = 1\)

Key concepts

Complex Numbers

Complex numbers are numbers that have both a real part and an imaginary part. They are usually written in the form \(z = a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit with the property \(i^2 = -1\). These numbers expand our ability to solve equations that do not have solutions within just the set of real numbers.

• Real Part: This is the \(a\) in \(a + bi\). It's a regular real number, like 2, -3, or 0.5.
• Imaginary Part: This is the \(bi\) part, where \(b\) is multiplied by \(i\). For example, if \(b = 4\), the imaginary part is \(4i\).

Understanding complex numbers becomes crucial when dealing with power expressions in the complex plane, as they help define complex functions and transformations. When we engage with complex numbers using the real and imaginary parts, mathematics can explore beyond the dimensions limited to real numbers alone.
They allow us to easily switch between different forms of expressions, particularly when incorporating trigonometric forms like in this exercise.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are essential in understanding the geometric representation of complex numbers. When applied to complex numbers, these functions help us describe the number in terms of its angle and magnitude. This representation is particularly beneficial when raising complex numbers to a power.Using Euler's formula, we can express a complex number in polar form:\[ z = r(\cos \theta + i \sin \theta) \]

• Cosine Function \(\cos \theta\): Represents the horizontal coordinate on the unit circle associated with an angle \(\theta\).
• Sine Function \(\sin \theta\): Represents the vertical coordinate on the unit circle for the same angle, \(\theta\).

For this exercise, the given complex expression was in trigonometric form already, \(\cos (3 \pi / 2) + i \sin (3 \pi / 2)\). It combines trigonometric functions to express rotations in the complex plane.
Understanding these functions as parts of complex numbers helps simplify calculations, especially through De Moivre's theorem, which will be applied for raising powers.

Powers of Complex Numbers

Raising a complex number to a power is made straightforward by De Moivre's theorem. This theorem states that for a complex number in polar form, \( z = r(\cos \theta + i \sin \theta) \), raising this number to the \(n\)th power, results in:\[ z^n = r^n(\cos(n \theta) + i \sin(n \theta)) \]This formula simplifies the process by reducing it to a recalculation of the angle and magnitude.

• Calculate New Angle: Multiply the original angle \(\theta\) by the power \(n\).
• Maintain Magnitude: If the magnitude \(r\) is 1, as it often is on the unit circle, then \(r^n\) simply equals 1.

In our case, the problem was solved using De Moivre’s theorem by multiplying the angle \(3 \pi / 2\) by 6, arriving at \(9 \pi\).
The resulting trigonometric calculation yielded \(\cos (9 \pi) = 1\) and \(\sin (9 \pi) = 0\). Thus, the power of the original complex number simplified beautifully to 1, making the process both elegant and powerful.