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To solve problems involving complex numbers, converting the number into its trigonometric form can be very helpful. The trigonometric form is expressed as \( r(\cos \theta + i\sin \theta) \), where \( r \) is the modulus (or magnitude) and \( \theta \) is the argument (or angle with the positive real axis).
For example, for the complex number \( i \), we have a real part of 0 and an imaginary part of 1. Calculating the modulus gives us \( r = \sqrt{0^2 + 1^2} = 1 \).
The argument \( \theta \) is \( \frac{\pi}{2} \) because \( i \) is purely imaginary and lies on the positive imaginary axis. So in trigonometric form, \( i \) is expressed as \( 1(\cos \frac{\pi}{2} + i\sin \frac{\pi}{2}) \). Understanding this form will make it easier to perform operations like finding roots.

De Moivre's Theorem is a powerful tool that simplifies the process of finding powers and roots of complex numbers in trigonometric form. It states that if \( z = r(\cos \theta + i\sin \theta) \), then for any integer \( n \), \( z^n = r^n(\cos(n\theta) + i\sin(n\theta)) \).
When finding roots, we slightly adjust this concept. To find cube roots, for example, we use \( n = 3 \), meaning each root is of the form \( r^{1/3}(\cos \frac{\theta + 2k\pi}{3} + i\sin \frac{\theta + 2k\pi}{3}) \) for different \( k \).
This means cycling through \( k = 0, 1, \) and \( 2 \) covers all possible cube roots, taking advantage of the periodic nature of sine and cosine functions to spread the roots evenly in a circular path around the origin.

Finding cube roots of a complex number involves using the trigonometric form and a methodical application of De Moivre’s Theorem. Each complex number has exactly \( n \) distinct \( n \)-th roots. For cube roots, this number is 3.
For the complex number \( i \), we calculate roots for \( k = 0, 1, \) and \( 2 \):

• When \( k = 0 \): The angle becomes \( \frac{\pi}{6} \), leading to the root \( \frac{\sqrt{3}}{2} + i\frac{1}{2} \).
• When \( k = 1 \): The angle becomes \( \frac{5\pi}{6} \), producing the root \( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \).
• When \( k = 2 \): The angle simplifies to \( \pi \), giving the root \( -i \).

These roots form an equilateral triangle in the complex plane, showing the symmetry and periodic nature of complex roots.

Graphing complex numbers in the complex plane turns each number into a point, or vector, originating from the origin and extending into a unique position defined by its modulus and argument.
For the cube roots of \( i \) we calculated:

• The root \( \frac{\sqrt{3}}{2} + i\frac{1}{2} \) is positioned in the first quadrant.
• The root \( -\frac{\sqrt{3}}{2} + i\frac{1}{2} \) is in the second quadrant.
• The root \( -i \) lies on the negative imaginary axis, pointing downward.

The complexes appear as vectors extending from the origin, illustrating the radial and angular attributes defined in trigonometric form. These visual representations on the complex plane often help reinforce the symmetry and spacing of the roots, giving a concrete grasp of their values and how they relate to each other.