Problem 86 - Precalculus Mathematics for Calculus

Question

Find the indicated roots, and graph the roots in the complex plane. The cube roots of \(1+i\)

Step-by-Step Solution

Answer

The cube roots of \(1+i\) are \( \sqrt[3]{\sqrt{2}}(\cos \frac{\pi}{12} + i\sin \frac{\pi}{12}) \), \( \sqrt[3]{\sqrt{2}}(\cos \frac{5\pi}{12} + i\sin \frac{5\pi}{12}) \), and \( \sqrt[3]{\sqrt{2}}(\cos \frac{9\pi}{12} + i\sin \frac{9\pi}{12}) \).

Step 1: Express the Complex Number in Polar Form

The given complex number is \( 1 + i \). To express it in polar form, we find the magnitude and argument. The magnitude is \( |1+i| = \sqrt{1^2 + 1^2} = \sqrt{2} \). The argument is \( \theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4} \). Thus, the polar form of \( 1+i \) is \( \sqrt{2}( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) \).

Step 2: Apply De Moivre's Theorem for Cube Roots

Using De Moivre's Theorem, the cube roots of a complex number \( r (\cos \theta + i \sin \theta) \) are given by \( \sqrt[3]{r} \left(\cos \frac{\theta + 2k\pi}{3} + i \sin \frac{\theta + 2k\pi}{3}\right) \) for \( k = 0, 1, 2 \). Here, \( r = \sqrt{2} \) and \( \theta = \frac{\pi}{4} \). The magnitude of the cube roots is \( \sqrt[3]{\sqrt{2}} \).

Step 3: Calculate Each Cube Root

For \( k=0 \), we get \( \sqrt[3]{\sqrt{2}} \left(\cos \frac{\pi}{12} + i \sin \frac{\pi}{12}\right) \). For \( k=1 \), we have \( \sqrt[3]{\sqrt{2}} \left(\cos \frac{\pi+2\pi}{12} + i \sin \frac{\pi+2\pi}{12}\right) \) or \( \sqrt[3]{\sqrt{2}} \left(\cos \frac{\pi}{12} + \frac{\pi}{6} + i \sin \frac{\pi}{2}\right) \). For \( k=2 \), it becomes \( \sqrt[3]{\sqrt{2}} \left(\cos \frac{\pi+4\pi}{12} + i \sin \frac{\pi+4\pi}{12}\right) \). Evaluate these trigonometric expressions to get the values of the roots.

Step 4: Plot the Roots in the Complex Plane

Calculate the exact values of the trigonometric expressions in Step 3 to get approximate Cartesian coordinates for each root. Plot these points on the complex plane. Each root corresponds to a point, and they should be equally spaced in a circular pattern, since they're roots of unity type.

Key Concepts

Polar Form

To grasp the concept of polar form for complex numbers, let's consider the complex number \(1 + i\). The idea is to represent this number in terms of its magnitude (or modulus) and direction angle (or argument). The polar form is very helpful for understanding the geometric representation of complex numbers.

Here's how you can convert a complex number to its polar form:

• **Magnitude:** This is the distance from the origin to the point \((x, y)\) in the complex plane. For \(1 + i\), the magnitude is \(|1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}\).
• **Argument:** This is the angle formed with the positive real axis. For \(1 + i\), the argument is \(\theta = \tan^{-1}(\frac{1}{1}) = \frac{\pi}{4}\).

Combining these, the polar form of \(1+i\) is \( \sqrt{2}( \cos\frac{\pi}{4} + i\sin\frac{\pi}{4}) \). This form is essential for simplifying the calculation of powers and roots of complex numbers.

Cube Roots

Finding the cube roots of a complex number involves knowing how to express the number in polar form and then manipulating it to find roots. This process stems from the properties of exponents and their geometric implications on the complex plane.

When we look for cube roots (or any nth roots), we essentially seek three numbers whose cube equals the original complex number. Here's what you need to do:

• Convert the number to its polar form, as previously explained.
• Apply the formula: each root is given by \( \sqrt[3]{r}(\cos\frac{\theta + 2k\pi}{3} + i \sin\frac{\theta + 2k\pi}{3}) \).
• In practice, compute each root by considering each value of \(k\) (e.g., \(k = 0, 1, 2\)).

This process lets you find all cube roots, which should lie equally spaced on an arc (circle segment) in the complex plane. Understanding this ensures you can visualize and verify the accuracy of your results.

De Moivre's Theorem

De Moivre's Theorem is a cornerstone in working with powers and roots of complex numbers, particularly in polar form. It states that for a complex number expressed as \(r(\cos\theta + i\sin\theta)\), the n-th power is given by \(r^n(\cos(n\theta) + i\sin(n\theta))\).

What makes De Moivre's Theorem exceptionally useful for roots is its adaptation for fractional powers:

• To find the nth roots, adjust the exponent as \(1/n\), which modifies the angle as \((\theta + 2k\pi)/n\).
• This adjustment accounts for the modulus and the angle, giving all possible nth roots geometrically spread around the origin.

In our example, applying De Moivre's Theorem provides the cube roots of \(1+i\) by ensuring they are evenly distributed across a circle in the complex plane, maintaining symmetry and periodicity.

Complex Plane

The complex plane is an elegant way to visualize complex numbers geometrically. It functions similarly to the Cartesian coordinate system but describes two-dimensional space in terms of real and imaginary components.

Visualizing complex numbers and their operations involves placing them on this plane:

• The horizontal axis (x-axis) is the line of real numbers.
• The vertical axis (y-axis) represents imaginary numbers.
• Each complex number \(x + yi\) translates to a unique point \((x, y)\) on the plane.

For the exercise of plotting cube roots of \(1+i\), you plot each root calculated separately to see their positions. They should form vertices of an equilateral triangle centered on the origin, thanks to the symmetry in the cube root distribution due to De Moivre's Theorem. This geometric perspective helps deepen understanding and offers visual confirmation of numerical results.