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Magazine Chapter 8: Problem 92
Solve the equation. $$z^{8}-i=0$$
Short Answer
Expert verified
The solutions are \( z_k = e^{i(\frac{\pi}{2} + 2k\pi)/8} \) for \( k = 0, 1, 2, \ldots, 7 \).
Step by step solution
01
Rewrite the Equation
The given equation is \( z^8 - i = 0 \). Start by adding \( i \) to both sides to isolate \( z^8 \) on one side: \( z^8 = i \).
02
Express the Right Side in Polar Form
The complex number \( i \) is purely imaginary and can be expressed in polar form as \( i = 1e^{i\frac{\pi}{2}} \). Recall that \( i \) is on the imaginary axis at a 90-degree angle (\( \frac{\pi}{2} \) radians) from the positive real axis.
03
Use De Moivre's Theorem
Apply De Moivre's theorem to find \( z \): If \( z^8 = re^{i \theta} \), then \( z = r^{1/8}e^{i(\theta + 2k\pi)/8} \) for \( k = 0, 1, 2, \ldots, 7 \). Here, \( r = 1 \) and \( \theta = \frac{\pi}{2} \).
04
Calculate the Magnitude of z
Since \( r = 1 \), the magnitude of \( z \) is \( 1^{1/8} = 1 \). So, all solutions will lie on the unit circle in the complex plane.
05
Find the Solutions
For each \( k \) from 0 to 7, calculate \( z_k = e^{i(\frac{\pi}{2} + 2k\pi)/8} \). These calculations give the eight solutions:- \( z_0 = e^{i\frac{\pi}{16}} \)- \( z_1 = e^{i\frac{5\pi}{16}} \)- \( z_2 = e^{i\frac{9\pi}{16}} \)- \( z_3 = e^{i\frac{13\pi}{16}} \)- \( z_4 = e^{i\frac{17\pi}{16}} \)- \( z_5 = e^{i\frac{21\pi}{16}} \)- \( z_6 = e^{i\frac{25\pi}{16}} \)- \( z_7 = e^{i\frac{29\pi}{16}} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
De Moivre's Theorem
De Moivre's Theorem is a powerful tool used to find powers and roots of complex numbers when they are expressed in polar form. The theorem states that
- if a complex number is given as \( z = re^{i\theta} \), then its \( n \)-th power can be found by \( z^n = r^n e^{in\theta} \).
- For the root finding, \( z^{1/n} = r^{1/n} e^{i(\theta + 2k\pi)/n} \) for \( k = 0, 1, 2, \cdots, n-1 \).
Polar Form
The polar form of a complex number simplifies the process of finding powers and roots. In polar form, a complex number \( z \) is expressed as \( re^{i\theta} \), where:
- \( r \) is the magnitude (or modulus) of the complex number,
- \( \theta \) is the argument (or angle) with the positive real axis.
Unit Circle
The unit circle is a key geometric concept that helps visualize complex numbers. It is a circle with a radius of one centered at the origin of the complex plane. Any complex number on the unit circle has a magnitude of 1.
When graphing the eight roots found in our example, it鈥檚 critical to realize that all these roots will lie on the unit circle since each has a modulus of 1.
When graphing the eight roots found in our example, it鈥檚 critical to realize that all these roots will lie on the unit circle since each has a modulus of 1.
- This means the solutions are evenly distributed around the circle.
- Each root represents an equal angular rotation from the previous root, specifically \( \frac{\pi}{4} \) radians apart in our case.
Roots of Complex Numbers
Finding the roots of complex numbers involves uncovering multiple solutions that are evenly spaced around a circle in the complex plane. The number of roots is equal to the degree of the root you're finding. For our exercise of solving \( z^8 - i = 0 \), we find eight roots because we are dealing with an eighth root.
- These roots emerge from applying De Moivre鈥檚 Theorem, plotting them systematically around a circle.
- Each root corresponds to a point where the complex power returns to the original magnitude and angle, completing a full rotation.
