Problem 70 - Precalculus with Limits

Question

Finding a Power of a Complex Number In Exercises \(65-80\) , 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 solution to the exercise is -32.

Step 1: Convert the Complex Number to Polar Form

The complex number \(1 - \sqrt{3} i\) can be written in polar form as \(r(\cos(\theta) + i\sin(\theta))\), where \(r = \sqrt{a^2 + b^2}\) and \(\theta = \arctan(\frac{b}{a})\). In this case, \(r = 2\) and \(\theta = - \frac{\pi}{3}\). So, the polar form is \(2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3}))\).

Step 2: Apply DeMoivre's theorem

DeMoivre's theorem states that \((r(\cos(\theta) + i\sin(\theta)))^n = r^n(\cos(n\theta) + i\sin(n\theta))\). Applying this theorem to the number \(2(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3}))\) cubed, we get \(8(\cos(-\pi) + i\sin(-\pi))\).

Step 3: Convert back to rectangular form and multiply by 4

Now, we can convert the result back to rectangular form. We know that \(\cos(-\pi) = -1\) and \(\sin(-\pi) = 0\), so we get \(8(-1) = -8\). Finally, although we are multiplying the number \(1 - \sqrt{3} i\) by itself three times, the entire expression is multiplied by 4 because of the factor in front of the parentheses. So, the answer is \(4 \times -8 = -32\).

Key Concepts

Complex Numbers

Complex numbers are a sophisticated concept in mathematics that expand the traditional understanding of numbers beyond the real number line. They are of the form
\( a + bi \), where \( a \) is the real part and \( bi \) is the imaginary part, with \( i \) denoting the square root of -1. This peculiar creation opens up new dimensions for calculation and representation of numbers, allowing for solutions to equations that would otherwise lack real solutions, such as \( x^2 + 1 = 0 \).

In relation to our exercise, the complex number given is \( 1 - \sqrt{3} i \). Here, \( 1 \) is the real part, and \( -\sqrt{3} i \) is the imaginary part. Operating with complex numbers, especially when raising them to a power, can be a challenge. But techniques like DeMoivre's Theorem greatly simplify these operations, particularly when complex numbers are expressed in their polar or trigonometric forms.

Polar Form of Complex Numbers

The polar form is a powerful approach to express complex numbers based on their magnitude and direction.
Polar form describes a complex number not by its position on the standard Cartesian coordinate plane, but by how far away it is from the origin (its magnitude, \( r \)) and by the angle it forms with the positive real axis (its argument, \( \theta \)). In formal terms, it's written as
\[ r(\cos(\theta) + i\sin(\theta)) \]
where \( r \) is the modulus of the complex number and is calculated using the Pythagorean theorem as \( r = \sqrt{a^2 + b^2} \), and \( \theta \), also known as the argument, is determined by \( \theta = \arctan(\frac{b}{a}) \).
In our exercise, we translated the complex number to its polar form using these definitions to harness the simplicity of DeMoivre's Theorem for computing powers of complex numbers.

Trigonometric Form of Complex Numbers

Trigonometric form is essentially another name for the polar form of complex numbers and uses trigonometry to represent complex numbers. This is particularly useful when dealing with powers and roots of complex numbers.
To put a complex number into trigonometric form, we find \( r \) and \( \theta \) as we do for the polar form. It's the same \( r(\cos(\theta) + i\sin(\theta)) \) format. Applied trigonometry simplifies complex multiplication and division, as angles can be added or subtracted and radii can be multiplied or divided.

DeMoivre's Theorem is a classic example of the utility of trigonometric form, stating that \( (r(\cos(\theta) + i\sin(\theta)))^n = r^n(\cos(n\theta) + i\sin(n\theta)) \), enabling the straightforward computation of powers as seen in our exercise. The initial polar form makes it easier to apply the theorem and to find the complex number raised to any integer power.