Problem 75 - College Trigonometry

Question

Find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. the six sixth roots of \(z=64\)

Step-by-Step Solution

Answer

The sixth roots in rectangular form are: 2, 1 + i√3, -1 + i√3, -2, -1 - i√3, 1 - i√3.

Step 1: Express in Polar Form

The complex number given is 64, which is a real number with no imaginary part. In polar form, any real number can be written as: \( r \text{cis} \theta \), where \( r \) is the magnitude and \( \theta \) is the argument. For 64, the magnitude \( r = 64 \) and \( \theta = 0 \) since it lies on the positive real axis. Hence, 64 can be written as \( 64 \text{cis} 0 \).

Step 2: Calculate Principal nth Root

To find the sixth roots, we use the formula for nth roots: \( z^{1/n} = r^{1/n} \text{cis} \left( \frac{\theta + 2k\pi}{n} \right) \) where \( k = 0, 1, 2, ..., n-1 \). For \( n = 6 \), \( r^{1/6} = 64^{1/6} = 2 \). The angles will be given by \( \frac{0 + 2k\pi}{6} \).

Step 3: Break Down Root Angles

Now calculate the angles for each \( k \):- For \( k = 0 \), angle is \( \frac{0}{6} = 0 \).- For \( k = 1 \), angle is \( \frac{2\pi}{6} = \frac{\pi}{3} \).- For \( k = 2 \), angle is \( \frac{4\pi}{6} = \frac{2\pi}{3} \).- For \( k = 3 \), angle is \( \frac{6\pi}{6} = \pi \).- For \( k = 4 \), angle is \( \frac{8\pi}{6} = \frac{4\pi}{3} \).- For \( k = 5 \), angle is \( \frac{10\pi}{6} = \frac{5\pi}{3} \).

Step 4: Express Sixth Roots in Polar Form

The sixth roots in polar form are:- \( 2 \text{cis} 0 \)- \( 2 \text{cis} \frac{\pi}{3} \)- \( 2 \text{cis} \frac{2\pi}{3} \)- \( 2 \text{cis} \pi \)- \( 2 \text{cis} \frac{4\pi}{3} \)- \( 2 \text{cis} \frac{5\pi}{3} \)

Step 5: Convert to Rectangular Form

The rectangular form is given by \( r(\cos(\theta) + i\sin(\theta)) \).- \( 2(\cos(0) + i\sin(0)) = 2 + 0i \)- \( 2(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 1 + i\sqrt{3} \)- \( 2(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3})) = -1 + i\sqrt{3} \)- \( 2(\cos(\pi) + i\sin(\pi)) = -2 + 0i \)- \( 2(\cos(\frac{4\pi}{3}) + i\sin(\frac{4\pi}{3})) = -1 - i\sqrt{3} \)- \( 2(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3})) = 1 - i\sqrt{3} \)

Key Concepts

Polar Form

Polar form is a way of expressing complex numbers that highlights their magnitude and angle from the positive real axis. It is particularly useful when dealing with complex numbers because it simplifies multiplication and division. In polar form, a complex number is written as:

• \( z = r \text{cis} \theta \)

Here, \( r \) is the magnitude or absolute value of the complex number, and \( \theta \) is the argument or angle. The expression \( \text{cis} \theta \) is shorthand for \( \cos \theta + i\sin \theta \). Calculating the polar form involves finding these two components:

• Magnitude (\( r \)): This is the distance of the complex number from the origin on the complex plane. For real numbers like 64, the magnitude is simply the absolute value of the number.
• Argument (\( \theta \)): This is the angle formed with the positive real axis. For 64, which lies on the real axis, \( \theta = 0 \) since it doesn't form any angle.

Using these, the polar form for the complex number 64 is expressed as \( 64 \text{cis} 0 \), emphasizing its magnitude and position on the real axis.

Rectangular Form

Rectangular form provides another way to express complex numbers, using both real and imaginary components. A number in rectangular form looks like:

• \( z = a + bi \)

Where \( a \) is the real part and \( bi \) is the imaginary part. This is the most intuitive form for many people, resembling traditional arithmetic. To convert from polar to rectangular form, you use the equation:

• \( r(\cos \theta + i\sin \theta) \)

For each sixth root of 64:

• root: Calculated by \( r = 2 \text{ (since } 64^{1/6} = 2) \)
• angles: Compute \( r(\cos(\theta) + i\sin(\theta)) \) for each \( \theta \).

For example, at \( \theta = 0 \), the complex number is expressed in rectangular form as \( 2 + 0i \). Other angles provide the complex forms such as \( 1 + i\sqrt{3} \) and \(-1 - i\sqrt{3} \), showing the versatility of moving between forms to better suit the context of the problem.

nth Roots

The nth root of a complex number involves finding a complex number that, when raised to the nth power, equals the original number. This process uses the nth root theorem, which states:

• \( z^{1/n} = r^{1/n} \text{cis} \left( \frac{\theta + 2k\pi}{n} \right) \)

Here, \( k \) ranges from 0 to \( n-1 \), providing the various "angular" solutions. The calculation:

• Compute the principal root by taking the nth root of the modulus: \( r^{1/n} \).
• Determine each angle by \( \frac{\theta + 2k\pi}{n} \), which evenly spaces roots around the complex plane.

In our example of finding the sixth roots of 64:

• Magnitude of Root: 64 has a sixth root magnitude of 2.
• Angles: Are calculated as \( \frac{2k\pi}{6} \) for \( k = 0 \) to 5, giving six distinct roots in polar form.

Each root presents a unique angle, displaying the beauty of complex numbers where the same result can fan out to occupy multiple orientations on the complex plane.

Complex Numbers

Complex numbers are fascinating mathematical entities comprising a real and an imaginary part. Written in the form \( a + bi \), they extend the concept of one-dimensional number lines to two-dimensional complex planes:

• Real Part (\( a \)): Represents the horizontal component on the complex plane.
• Imaginary Part (\( bi \)): Represents the vertical component, where \( i \) is the imaginary unit satisfying \( i^2 = -1 \).

Understanding complex numbers is crucial for exploring advanced mathematics, especially in fields like electrical engineering and quantum physics. Solving for their roots, as in the exercise given, broadens comprehension of how complex numbers interact: spreading out into various directions in the complex plane and offering a rich tapestry of solutions. Each route, be it polar or rectangular, allows different insights, demonstrating the interplay between magnitude and direction. Moreover, complex numbers connect greatly diverse areas of math, bridging concepts from algebra, trigonometry, to even calculus.