Chapter 11: Problem 71
Find the indicated complex roots. Express your answers in polar form and then convert them into rectangular form. the three cube roots of \(z=i\)
Short Answer
Expert verified
The cube roots are \(\frac{\sqrt{3}}{2} + \frac{1}{2}i\), \(-\frac{\sqrt{3}}{2} + \frac{1}{2}i\), and \(-i\).
Step by step solution
01
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 \(\text{cis } \theta = \cos \theta + i \sin \theta\).
02
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 \(r^{1/3} = 1\).
03
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 \(1 \cdot \text{cis} \left(\frac{\pi/2 + 2(2)\pi}{3}\right) = \text{cis} \frac{3\pi/2}\).
04
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 \(\cos \frac{3\pi}{2} + i\sin \frac{3\pi}{2} = 0 - i = -i\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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 \(1 \text{cis} \frac{\pi}{2}\).
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:
Converting from polar to rectangular form often involves using trigonometric identities:
- \(\cos \theta = \frac{\text{adjacent side}}{\text{hypotenuse}}\)
- \(\sin \theta = \frac{\text{opposite side}}{\text{hypotenuse}}\)
- \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\)
- \(\sin \frac{\pi}{6} = \frac{1}{2}\)
- \(\frac{\sqrt{3}}{2} + \frac{1}{2}i\)
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:
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 \(k = 0, 1, 2\).
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:
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 \(\text{cis} \frac{3\pi}{2}\)
- \(\text{cis} \frac{\pi}{6} = \frac{\sqrt{3}}{2} + \frac{1}{2}i\)
- \(\text{cis} \frac{5\pi}{6} = -\frac{\sqrt{3}}{2} + \frac{1}{2}i\)
- \(\text{cis} \frac{3\pi}{2} = -i\)