Problem 70 - College Trigonometry

Question

Find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. the three cube roots of \(z=-125\)

Step-by-Step Solution

Answer

The cube roots of \(-125\) are \(\frac{5}{2} + i\frac{5\sqrt{3}}{2}\), \(-5\), and \(\frac{5}{2} - i\frac{5\sqrt{3}}{2}\).

Step 1: Express the Complex Number in Polar Form

The number \(-125\) is a real number located on the negative real axis. To express it in polar form, we find the modulus and the argument. The modulus is the absolute value, \( |z| = 125 \). The argument is the angle, which is \(\pi\) radians (or 180 degrees) because it is on the negative real axis. Thus, in polar form, \( z = 125\text{cis}(\pi) \), where \(\text{cis}(x) = \cos(x) + i\sin(x) \).

Step 2: Use De Moivre's Theorem for the Cube Roots

To find the cube roots of a complex number, we apply De Moivre's Theorem. For a complex number \(z = r\text{cis}(\theta)\), its \(n^{th}\) roots are given by \(z_k = r^{1/n}\text{cis}\left(\frac{\theta + 2k\pi}{n}\right)\) for \(k = 0, 1, ..., n-1\). Here, \(n=3\), so each root \(z_k\) is \(5\text{cis}\left(\frac{\pi + 2k\pi}{3}\right)\) since \(125^{1/3} = 5\).

Step 3: Calculate the Three Cube Roots in Polar Form

Substituting \(k = 0, 1, 2\) into the formula, we find:- For \(k = 0\), \(z_0 = 5\text{cis}\left(\frac{\pi}{3}\right)\).- For \(k = 1\), \(z_1 = 5\text{cis}\left(\frac{\pi}{3} + \frac{2\pi}{3}\right) = 5\text{cis}(\pi)\).- For \(k = 2\), \(z_2 = 5\text{cis}\left(\frac{\pi}{3} + \frac{4\pi}{3}\right) = 5\text{cis}\left(\frac{5\pi}{3}\right)\).

Step 4: Convert the Polar Form to Rectangular Form

To convert from polar to rectangular form, use the identity \( \text{cis}(\theta) = \cos(\theta) + i\sin(\theta) \). Calculate each:- \(z_0 = 5(\cos(\frac{\pi}{3}) + i\sin(\frac{\pi}{3})) = 5(\frac{1}{2} + i\frac{\sqrt{3}}{2}) = \frac{5}{2} + i\frac{5\sqrt{3}}{2}\).- \(z_1 = 5(\cos(\pi) + i\sin(\pi)) = 5(-1) = -5\).- \(z_2 = 5(\cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3})) = 5(\frac{1}{2} - i\frac{\sqrt{3}}{2}) = \frac{5}{2} - i\frac{5\sqrt{3}}{2}\).

Key Concepts

Polar Form

Understanding the polar form of a complex number is essential in grasping complex roots. The polar form represents a complex number as a point in the complex plane using a magnitude (or modulus) and an angle (or argument).
For a complex number like \(z = -125\), we convert it to polar form by finding two key components:

• Modulus: This is the distance from the origin to the point, which is simply the absolute value. Here, the modulus is \(|z| = 125\).
• Argument: This is the angle formed with the positive real axis. Since \(-125\) is on the negative real axis, the argument is \(\pi\) radians (or 180 degrees).

Once these components are identified, we express the complex number in polar form as \(|z|\text{cis}(\text{arg}) = 125\text{cis}(\pi)\), where \(\text{cis}(x) = \cos(x) + i\sin(x)\).

Rectangular Form

When solving complex numbers, transforming from polar to rectangular form makes computations easier, enhancing understanding of their geometric representation.
In rectangular form, a complex number is expressed as \(x + yi\), where \(x\) is the real part and \(yi\) represents the imaginary part.To convert a polar form like \(5\text{cis}(\theta)\) to rectangular form:

• Real Part (x): \(x = 5\cos(\theta)\)
• Imaginary Part (y): \(y = 5\sin(\theta)\)

Using trigonometric identities, such as \(\cos(\theta)\text{ and }\sin(\theta)\), allows for easy conversion between polar and rectangular forms, deepening the understanding of the complex number’s structure.

De Moivre's Theorem

De Moivre's Theorem acts as a bridge to link complex numbers in polar form to their powers and roots, particularly useful for calculations involving cube roots and higher.
The theorem states that for a complex number \(z = r\text{cis}(\theta)\), the \(n^{th}\) root is expressed as:\[z_k = r^{1/n}\text{cis}\left(\frac{\theta + 2k\pi}{n}\right)\] for \(k = 0, 1, \ldots, n-1\).When we need to find the cube roots of \(z = -125\), we use \(n=3\) to determine each root. This approach simplifies complex calculations by breaking them into manageable angles and magnitudes.

Cube Roots

Cube roots in the context of complex numbers involve finding three distinct values that, when raised to the third power, return the original number.
Applying De Moivre’s Theorem helps to uncover these roots efficiently. Taking \(z = -125\) as an example, we need to determine each root:

• The modulus of each root is \(125^{1/3} = 5\).
• The angular arrangement involves dividing \(360^{\circ}\) (or \(2\pi\) radians) into three equal parts, corresponding to the \(n = 3\).

Each root is derived by adjusting the initial argument \(\pi\) by \(\frac{2\pi k}{3}\) for \(k = 0, 1, 2\). These calculated roots are then translated back into their respective rectangular forms for practical applications.

Complex Number Conversion

Complex number conversion is a vital skill, allowing shifts between forms to facilitate a variety of mathematical and practical applications.
In our case, the conversion from polar to rectangular form involves applying trigonometric functions to accurately depict complex numbers on the plane. For instance, turning \(5\text{cis}(\theta)\) back into \(x + yi\) form.

• Start with known angle and magnitude.
• Use \(\cos(\theta)\text{ for the real part, and } \sin(\theta)\text{ for the imaginary part.}\)

Such conversions not only make complex numbers tangible but also simplify operations like addition, subtraction, and finding roots, essential for deeper mathematical exploration.