Problem 71 - 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 \(z=i\)

Step-by-Step Solution

Answer

The cube roots are \(\frac{\sqrt{3}}{2} + \frac{1}{2}i\), \(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\), and \(-i\).

Step 1: Convert to Polar Form

The complex number \(z = i\) can be seen as \(0 + 1i\) in rectangular form. To convert it to polar form, we need to find the modulus and the argument. The modulus \(|z|\) is computed as \(|z| = \sqrt{0^2 + 1^2} = 1\). The argument \(\theta\) is \(\frac{\pi}{2}\) since the angle required to get from the positive x-axis to the positive imaginary unit is \(90^\circ\). Thus, the polar form is \(z = 1 \text{cis} \frac{\pi}{2}\), where \(\text{cis } \theta = \cos \theta + i \sin \theta\).

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

According to the De Moivre's Theorem, the cube roots of \(z\) in polar form \(r \text{cis } \theta\) are given by \(r^{1/3} \text{cis} \left(\frac{\theta + 2k\pi}{3}\right)\) for \(k = 0, 1, 2\). Here, \(r = 1\), hence \(r^{1/3} = 1\).

Step 3: Compute the Cube Roots

For \(k = 0\), the root is given by \(1 \cdot \text{cis} \left(\frac{\pi/2 + 2(0)\pi}{3}\right) = \text{cis} \frac{\pi}{6}\).For \(k = 1\), the root is \(1 \cdot \text{cis} \left(\frac{\pi/2 + 2(1)\pi}{3}\right) = \text{cis} \frac{5\pi}{6}\). For \(k = 2\), the root is \(1 \cdot \text{cis} \left(\frac{\pi/2 + 2(2)\pi}{3}\right) = \text{cis} \frac{3\pi/2}\).

Step 4: Convert Roots to Rectangular Form

The first root \(\text{cis } \frac{\pi}{6}\) converts to rectangular form as \(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i\).The second root \(\text{cis } \frac{5\pi}{6}\) converts to \(\cos \frac{5\pi}{6} + i\sin \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i\).The third root \(\text{cis } \frac{3\pi}{2}\) converts to \(\cos \frac{3\pi}{2} + i\sin \frac{3\pi}{2} = 0 - i = -i\).

Key Concepts

Polar Form

The polar form of a complex number is a way of expressing it using the modulus and the argument. It is often represented as \(z = r \text{cis} \theta \), where \(r\) is the modulus and \(\theta\) is the argument.

For the complex number \( z = i \), we can represent it in rectangular form as \( 0 + 1i \), which means the real part is 0 and the imaginary part is 1.

To convert this into polar form, we first calculate the modulus \(|z|\). The modulus is the distance from the origin to the point in the complex plane and is given by the formula \(|z| = \sqrt{x^2 + y^2}\). For \(z = i\), this becomes \(|z| = \sqrt{0^2 + 1^2} = 1\). The argument \(\theta\) is the angle from the positive x-axis to the line representing the complex number. For \(z = i\), \(\theta = \frac{\pi}{2}\) or 90 degrees because it points directly upwards on the imaginary axis.

• The polar form of \(z = i\) is thus \(1 \text{cis} \frac{\pi}{2}\).

This form is extremely useful when performing operations like multiplication or finding roots of complex numbers.

Rectangular Form

Rectangular form is another method of expressing complex numbers, written as \(a + bi\). This format clearly shows the real component \(a\) and the imaginary component \(b\) of the complex number \(z\).

Converting from polar to rectangular form often involves using trigonometric identities:

• \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
• \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)

Consider a complex number in its polar form: \(1 \text{cis} \frac{\pi}{6}\). It can be converted back to rectangular form like this:

• \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\)
• \(\sin \frac{\pi}{6} = \frac{1}{2}\)

Combine both to form the rectangular expression:

• \(\frac{\sqrt{3}}{2} + \frac{1}{2}i\)

Understanding both forms allows for flexible use of complex numbers in analysis and applications.

De Moivre's Theorem

De Moivre's Theorem is a fundamental principle when dealing with powers and roots of complex numbers. It states that for any complex number in polar form \((r \text{cis } \theta)\) and any integer \(n\), the equation is given by: \[(r \text{cis } \theta)^n = r^n \text{cis } (n\theta)\]

Use it to find roots or powers of complex numbers more swiftly.

For example, finding the cube roots of a complex number like \(z = i\) involves turning \(i\) into polar form \((1 \text{cis } \frac{\pi}{2})\) and then using De Moivre's Theorem to find the roots:

• \(r = 1\) means \(r^{1/3} = 1\)
• \(\theta = \frac{\pi}{2}\)
• Cubic roots are \(1 \text{cis } \left(\frac{\frac{\pi}{2} + 2k\pi}{3}\right),\)
where \(k = 0, 1, 2\).

It simplifies the otherwise cumbersome process of figuring out roots, especially in practical applications.

Cube Roots

Finding the cube roots of a complex number means solving an equation of the form \(z^3 = i\).

Using the polar form and De Moivre's Theorem as described earlier, you can easily determine the cube roots. Specifically, with \(k = 0, 1, 2\), each \(k\) gives a different root angle:

• \(k=0\) provides \(\text{cis} \frac{\pi}{6}\)
• \(k=1\) gives \(\text{cis} \frac{5\pi}{6}\)
• \(k=2\) results in \(\text{cis} \frac{3\pi}{2}\)

These directly translate to rectangular forms as:

• \(\text{cis} \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i\)
• \(\text{cis} \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i\)
• \(\text{cis} \frac{3\pi}{2} = -i\)

Cube roots often appear in electrical engineering and physics, representing periodic functions and waveforms.