Problem 70 - Precalculus with Limits

Question

In Exercises 67-82, use DeMoivre's Theorem to find the indicated power of the complex number. Write the result in standard form. $$ 4(1-\sqrt{3} i)^{3} $$

Step-by-Step Solution

Answer

The result is -32.

Step 1: Convert to Polar Form

A complex number can be expressed in the standard form \(a + bi\) or in polar form \(r(\cos θ + i \sin θ)\). For \(1-\sqrt{3}i\), \(a=1\) and \(b=-\sqrt{3}\). The magnitude \(r\) is \(\sqrt{a^2 + b^2} = 2\) and the argument \(θ\) is \(\tan^{-1}(-\sqrt{3}/1) = -\pi/3\). Therefore, the polar form is \(2(\cos(-\pi/3) + i \sin(-\pi/3))\).

Step 2: Apply DeMoivre's Theorem

DeMoivre's theorem says \((\cos θ + i \sin θ)^n = \cos(nθ) + i \sin(nθ)\). Here, \(n=3\), so the cube of the complex number in polar form is \(2^3(\cos(3*(-\pi/3)) + i \sin(3*(-\pi/3))) = 8(\cos\pi + i \sin \pi)\).

Step 3: Convert Back to Standard Form

\(\cos\pi = -1\) and \(\sin\pi = 0\), so the standard form of the complex number's cube is \(8(-1 + 0i) = -8\).

Step 4: Multiply by 4

Multiply the result by 4 to get \(4(-8) = -32\).

Key Concepts

Complex Numbers

Complex numbers are fascinating and extend our understanding of numbers beyond the real number line. They include a real part and an imaginary part, typically written as \(a + bi\), where \(a\) is the real part and \(b\) is the imaginary part. The imaginary part involves the imaginary unit, \(i\), defined as \(i^2 = -1\). This definition allows complex numbers to capture new dimensions beyond traditional numbers.
Understanding complex numbers involves recognizing how they can visualize quantities that lie outside the realm of real numbers. They provide solutions to equations that do not have real solutions, particularly quadratic equations with negative discriminants.

• The real part, \(a\), represents a point on the horizontal axis of the complex plane.
• The imaginary part, \(b\), represents a point on the vertical axis of the complex plane.
• When combined as \(a + bi\), they locate a specific point or vector in this two-dimensional space, known as the complex plane.

By understanding how each part contributes to the larger number, students can solve more advanced equations like the one in our exercise, applying important theorems such as DeMoivre's Theorem, which extends the calculation of powers and roots of complex numbers.

Polar Form

The polar form of complex numbers is another way to represent them, highlighting their magnitude and direction. Rather than using the rectangular form \(a + bi\), the polar form uses the magnitude \(r\) and angle \(θ\).
The magnitude (or modulus) \(r\) is the distance from the origin to the point in the complex plane, calculated as \(\sqrt{a^2 + b^2}\). The angle (or argument) \(θ\) is the direction of that point from the positive real axis, often calculated using \(\tan^{-1}(b/a)\).

• The conversion to polar form is useful for multiplication, division, and finding powers and roots of complex numbers.
• It simplifies complicated arithmetic operations into addition and multiplication because angles can be easily added or subtracted.
• The polar form is expressed as \(r(\cos θ + i \sin θ)\), making it ideal for applying DeMoivre's Theorem.

By understanding the polar form, we gain versatile tools for manipulating complex numbers, as demonstrated when finding higher powers of numbers efficiently, as in the exercise.

Standard Form

Standard form is the most common way of writing complex numbers, as \(a + bi\). This form allows for straightforward addition and subtraction but can be cumbersome for multiplication and division, especially when dealing with powers.
Returning to the standard form from computations done in polar form often involves using trigonometric values like \(\cos θ\) and \(\sin θ\). In the final step of our problem, the complex number was converted back from its polar power form to standard form.

• To get back to standard form from polar power form, one must evaluate the trigonometric values \(\cos nθ\) and \(\sin nθ\).
• The results create an ordinary expression without reliance on imaginary unit calculations.
• For this problem, the simplified expression demonstrated how the complex number's cube simplifies to a very straightforward real number.

Using the standard form, finding sums or differences becomes tactical, and expressing end results becomes clear and recognizable for further mathematical operations.