How to Find the 6th Root of 729i: A Comprehensive Guide to Complex Numbers and Roots

Understanding the complex number system and its operations, such as finding roots, is fundamental in many areas of mathematics, engineering, and physics. In this article, we explore the process of determining the 6th roots of 729i using Euler's formula and the properties of complex numbers.

Understanding Complex Numbers

Complex numbers are numbers of the form (a bi), where (a) is the real part and (b) is the imaginary part. The imaginary unit (i) is defined as (i sqrt{-1}). In this guide, we specifically focus on the complex number (729i), where the real part is 0 and the imaginary part is 729.

Euler's Formula and De Moivre's Theorem

To find the 6th roots of a complex number, we use Euler's formula and De Moivre's theorem. Euler's formula states that (e^{ix} cos x i sin x), where (i) is the imaginary unit. De Moivre's theorem extends this to powers of complex numbers, stating that for any real number (n), ((cos theta i sin theta)^n cos(ntheta) i sin(ntheta)).

Step-by-Step Solution: Finding the 6th Roots of 729i

Given the complex number (z 729i), we can express it in polar form. The modulus (or absolute value) of (729i) is (|729i| 729), and its argument (angle) is (frac{pi}{2}) radians, since it lies on the positive imaginary axis.

Expression in Polar Form

Using the polar form, we write:

(z 729 left( cos left( frac{pi}{2} right) i sin left( frac{pi}{2} right) right))

Applying De Moivre's Theorem

To find the 6th roots, we use De Moivre's theorem, which states that the (n)th roots of a complex number (z) in polar form (r(cos theta i sin theta)) are given by:

(w_k sqrt[n]{r} left( cos left( frac{theta 2kpi}{n} right) i sin left( frac{theta 2kpi}{n} right) right))

where (k 0, 1, 2, 3, 4, 5).

Calculating the 6th Roots

For (z 729i), (r 729) and (theta frac{pi}{2}). The 6th root of 729 is (sqrt[6]{729} 3). Thus, the 6th roots are:

(w_k 3 left( cos left( frac{frac{pi}{2} 2kpi}{6} right) i sin left( frac{frac{pi}{2} 2kpi}{6} right) right))

For (k 0, 1, 2, 3, 4, 5), we calculate the roots as follows:

For (k 0):

(w_0 3 left( cos left( frac{pi/2}{6} right) i sin left( frac{pi/2}{6} right) right))

( 3 left( cos left( frac{pi}{12} right) i sin left( frac{pi}{12} right) right))

For (k 1):

(w_1 3 left( cos left( frac{5pi/2}{6} right) i sin left( frac{5pi/2}{6} right) right))

( 3 left( cos left( frac{5pi}{12} right) i sin left( frac{5pi}{12} right) right))

For (k 2):

(w_2 3 left( cos left( frac{3pi/2}{6} right) i sin left( frac{3pi/2}{6} right) right))

( 3 left( cos left( frac{pi}{2} right) i sin left( frac{pi}{2} right) right))

( 3i)

For (k 3):

(w_3 3 left( cos left( frac{13pi/2}{6} right) i sin left( frac{13pi/2}{6} right) right))

( 3 left( cos left( pi frac{pi}{12} right) i sin left( pi frac{pi}{12} right) right))

( 3 left( -cos left( frac{pi}{12} right) - i sin left( frac{pi}{12} right) right))

For (k 4):

(w_4 3 left( cos left( frac{17pi/2}{6} right) i sin left( frac{17pi/2}{6} right) right))

( 3 left( cos left( pi frac{5pi}{12} right) i sin left( pi frac{5pi}{12} right) right))

( 3 left( -cos left( frac{5pi}{12} right) - i sin left( frac{5pi}{12} right) right))

For (k 5):

(w_5 3 left( cos left( frac{7pi/2}{6} right) i sin left( frac{7pi/2}{6} right) right))

( 3 left( cos left( 2pi - frac{pi}{4} right) i sin left( 2pi - frac{pi}{4} right) right))

( 3 left( cos left( frac{pi}{4} right) - i sin left( frac{pi}{4} right) right))

( 3 left( frac{sqrt{2}}{2} - frac{i sqrt{2}}{2} right))

( frac{3sqrt{2}}{2} - frac{3isqrt{2}}{2})

In summary, the 6th roots of (729i) are:

(w_0 3 left( cos frac{pi}{12} i sin frac{pi}{12} right))

(w_1 3 left( cos frac{5pi}{12} i sin frac{5pi}{12} right))

(w_2 3i)

(w_3 3 left( -cos frac{pi}{12} - i sin frac{pi}{12} right))

(w_4 3 left( -cos frac{5pi}{12} - i sin frac{5pi}{12} right))

(w_5 frac{3sqrt{2}}{2} - frac{3isqrt{2}}{2})

Conclusion

Understanding the process of finding roots of complex numbers is crucial for various scientific and engineering applications. By using Euler's formula and De Moivre's theorem, we can systematically determine the 6th roots of any complex number, as demonstrated in the step-by-step example above.