Find All Complex Cube Roots Of 4+i. Give Your Answers In A+bi Form, Separated By Commas.
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Introduction to Complex Cube Roots
In the fascinating realm of complex numbers, finding roots of complex expressions is a fundamental task that combines algebra, geometry, and trigonometry. Among these, computing cube roots of complex numbers is particularly interesting because each complex number, except zero, has exactly three cube roots in the complex plane. These roots are evenly spaced in argument (angle) and can be expressed in the form a + bi, where a and b are real numbers.
Today, we focus on finding all the cube roots of a specific complex number: 4 + i. The goal is to determine all solutions to the equation:
\[ z^3 = 4 + i \]
and express each root in the standard a + bi form.
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Understanding the Problem
Before diving into calculations, it's important to understand the key concepts involved:
- Complex numbers are expressed as a + bi, where a is the real part, and b is the imaginary part.
- Modulus (r) of a complex number \( z = a + bi \) is given by:
\[ r = |z| = \sqrt{a^2 + b^2} \]
- Argument (\(\theta\)) is the angle between the positive real axis and the line segment from the origin to the point (a, b):
\[ \theta = \arg(z) = \arctan\left(\frac{b}{a}\right) \]
- De Moivre's Theorem states that for a complex number in polar form \( z = r(\cos \theta + i \sin \theta) \), its \( n \)-th roots are given by:
\[ z_k = r^{1/n} \left( \cos \left(\frac{\theta + 2k\pi}{n}\right) + i \sin \left(\frac{\theta + 2k\pi}{n}\right) \right) \]
for \( k = 0, 1, 2, ..., n-1 \).
Applying these principles, our task involves converting 4 + i into polar form, determining its modulus and argument, and then applying De Moivre's theorem to find all cube roots.
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Step 1: Convert 4 + i to Polar Form
The first step is to compute the modulus \( r \) and argument \( \theta \).
Calculating the Modulus \( r \)
\[ r = \sqrt{(4)^2 + (1)^2} = \sqrt{16 + 1} = \sqrt{17} \]
Calculating the Argument \( \theta \)
\[ \theta = \arctan \left( \frac{1}{4} \right) \]
Since \( \frac{1}{4} \) is a small positive number, \( \theta \) is approximately:
\[ \theta \approx \arctan(0.25) \approx 0.244978 \text{ radians} \]
Expressed in degrees, this is roughly 14.04°, but for our calculations, we'll retain radians for precision.
Polar form of 4 + i
\[ 4 + i = r (\cos \theta + i \sin \theta) = \sqrt{17} \left( \cos 0.244978 + i \sin 0.244978 \right) \]
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Step 2: Apply De Moivre's Theorem to Find Cube Roots
The general formula for the cube roots of a complex number \( z = r (\cos \theta + i \sin \theta) \) is:
\[ z_k = r^{1/3} \left( \cos \left( \frac{\theta + 2k\pi}{3} \right) + i \sin \left( \frac{\theta + 2k\pi}{3} \right) \right) \]
for \( k = 0, 1, 2 \).
Compute \( r^{1/3} \)
\[ r^{1/3} = (\sqrt{17})^{1/3} = 17^{1/6} \]
which is approximately:
\[ 17^{1/6} \approx e^{\frac{1}{6} \ln 17} \]
Calculating:
\[ \ln 17 \approx 2.8332 \]
\[ \frac{1}{6} \times 2.8332 \approx 0.4722 \]
\[ e^{0.4722} \approx 1.603 \]
Thus:
\[ r^{1/3} \approx 1.603 \]
Calculate the three arguments
For each root \( z_k \), the argument is:
\[ \theta_k = \frac{\theta + 2k\pi}{3} \]
where \( k = 0, 1, 2 \).
- For \( k = 0 \):
\[ \theta_0 = \frac{0.244978 + 0}{3} = 0.08166 \text{ radians} \]
- For \( k = 1 \):
\[ \theta_1 = \frac{0.244978 + 2\pi}{3} = \frac{0.244978 + 6.283185}{3} \approx \frac{6.52816}{3} \approx 2.17605 \text{ radians} \]
- For \( k = 2 \):
\[ \theta_2 = \frac{0.244978 + 4\pi}{3} = \frac{0.244978 + 12.56637}{3} \approx \frac{12.81135}{3} \approx 4.27045 \text{ radians} \]
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Step 3: Write the Roots in a + bi Form
Now, for each \( k \), compute:
\[ zk = r^{1/3} (\cos \thetak + i \sin \theta_k) \]
with \( r^{1/3} \approx 1.603 \).
Root \( z_0 \) (when \( k=0 \))
\[ z_0 \approx 1.603 (\cos 0.08166 + i \sin 0.08166) \]
Calculating:
\[ \cos 0.08166 \approx 0.9967 \]
\[ \sin 0.08166 \approx 0.0816 \]
Therefore:
\[ z_0 \approx 1.603 \times 0.9967 + i \times 1.603 \times 0.0816 \]
\[ \approx 1.597 + 0.131 \]
Root \( z_1 \) (when \( k=1 \))
\[ z_1 \approx 1.603 (\cos 2.17605 + i \sin 2.17605) \]
Calculating:
\[ \cos 2.17605 \approx -0.573 \]
\[ \sin 2.17605 \approx 0.819 \]
Thus:
\[ z_1 \approx 1.603 \times (-0.573) + i \times 1.603 \times 0.819 \]
\[ \approx -0.918 + 1.312 \]
Root \( z_2 \) (when \( k=2 \))
\[ z_2 \approx 1.603 (\cos 4.27045 + i \sin 4.27045) \]
Calculating:
\[ \cos 4.27045 \approx -0.425 \]
\[ \sin 4.27045 \approx -0.905 \]
Therefore:
\[ z_2 \approx 1.603 \times (-0.425) + i \times 1.603 \times (-0.905) \]
\[ \approx -0.681 - 1.45 \]
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Final Answers in a + bi Form
Summarizing the roots:
- \( z_0 \approx 1.597 + 0.131i \)
- \( z_1 \approx -0.918 + 1.312i \)
- \( z_2 \approx -0.681 - 1.45i \)
Expressed precisely with rounded approximations, the three cube roots of \( 4 + i \) are:
Approximately:
\[ 1.597 + 0.131i, \, -0.918 + 1.312i, \, -0.681 - 1.45i \]
Exact forms involve radicals and trigonometric functions:
\[
z_k = 17^{1/6} \left( \cos \left( \frac{\arctan(1/4) + 2k\pi}{3} \right) + i \sin \left( \frac{\arctan(1/4) + 2k\pi}{3} \right) \right), \quad k=0,1,2
\]
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Conclusion
Finding all complex cube roots of a given complex number requires converting the number into polar form, applying De Moivre's theorem, and converting back to
