Lesson How to take a root of a complex number

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How to take a root of a complex number


Let n be a positive integer >= 2. The n-th root of a complex number w=a+bi is the complex number z=c+di such as z%5En+=+w.
Operation of extracting the root of the complex number is the inverse of raising a complex number to a power.
In the lesson Raising a complex number to an integer power in this module we derived De Moivre's formula



for the complex number presented in the trigonometric form, where r is its modulus and alpha is its argument.
Based on this formula, one can expect that the n-th root of the complex number r%28cos%28alpha%29+%2B+red%28i%29%2Asin%28alpha%29%29 is equal to root%28n%2Cr%29%2A%28cos%28alpha%2Fn%29%2Bred%28i%29%2Asin%28alpha%2Fn%29%29.
This is true that the complex number root%28n%2Cr%29%2A%28cos%28alpha%2Fn%29%2Bred%28i%29%2Asin%28alpha%2Fn%29%29, raised to degree n, is equal to r%28cos%28alpha%29+%2B+red%28i%29%2Asin%28alpha%29%29.
But this number is not unique: there are other complex numbers that have the same property.

Namely, numbers , k=1, 2, ...,n-1 are n-1 other distinct complex numbers that raised to degree n produce the same number r%28cos%28alpha%29+%2B+red%28i%29%2Asin%28alpha%29%29.

This follows directly from the De Moivre's formula.

Thus, the n-th root of the complex number r%28cos%28alpha%29+%2B+red%28i%29%2Asin%28alpha%29%29 has n distinct values , k = 0, 1, ...,n-1.
They all have the same modulus root%28n%2Cr%29 (positive real value) and n distinct argument values %28alpha+%2B+2k%2Api%29%2Fn, k = 0, 1, ...,n-1.
All n-roots are located on the circle of radius root%28n%2Cr%29 in the complex plane.
Also, since the argument of each successive n-th root is 2%2Api%2Fn apart the previous argument value, you see that n-th roots are equally spaced in this circle.

Examples

1) Square root of the complex number 1 (actually, this is the real number) has two values: 1 and -1.      
Surely, you know it well from your experience with real numbers (even with integer numbers).
In the frame of explanations given above, the number 1 has the modulus r=1 and the argument
alpha=0,2%2Api, 4%2Api,....
Taking the square root of 1, you have the modulus sqrt%281%29=1 (positive value) and two argument values
alpha=0 and alpha=2%2Api%2F2=pi. This gives two values of the square root of the complex number 1: 1 and -1,
exactly as for real numbers.
Figure 1 at the right illustrates how -1 arises as a result of manipulating the argument.


Figure 1. Square roots
    of the complex number 1



2) Square root of the complex number -1 (of the negative unit) has two values: i and -i.
It was explained in the lesson Complex numbers and arithmetical operations in this module, and                  
it is how the complex number i was introduced.
In the complex plane, the number -1 has the modulus r=1 and the argument
alpha=pi,pi%2B2%2Api=3%2Api, ....
Taking the square root of -1, you have the modulus sqrt%281%29=1 (positive value) and two argument values
alpha=pi%2F2 and alpha=3%2Api%2F2. This gives two values of the square root of the complex number -1: i and -i,
exactly as expected.
Figure 2 at the right illustrates how i and -i arise as a result of manipulating the argument.


Figure 2. Square roots
    of the complex number -1



3) Cube roots of a complex number 1.
In the plane of complex numbers, 1 has the modulus r=1 and the argument
values alpha=0,2%2Api, 4%2Api,....
Taking the cube root of the complex number 1, you have the modulus sqrt%281%29=1 (positive value) and                 
three argument values alpha=0, alpha=2%2Api%2F3 and alpha=4%2Api%2F3.
This gives three values of the cube root of the complex number 1: 1,
and .
Figure 3 shows complex (real) number 1 in red and two other roots in blue.
Note that these tree points are vertices of the equilateral triangle.


Figure 3. Cube roots
    of the complex number 1



4) 4-th roots of a complex number 1.
In the plane of complex numbers, 1 has the modulus r=1 and the argument
values alpha=0,2%2Api, 4%2Api, 6%2Api,....
Taking the 4-th root of the complex number 1, you have the modulus sqrt%281%29=1 (positive value) and                 
four argument values alpha=0, alpha=2%2Api%2F4=pi%2F2, alpha=%282%2Api%2B2%2Api%29%2F4=pi and alpha=%282%2Api%2B4%2Api%29%2F4=3%2Api%2F2.
This gives four values of the 4-th root of the complex number 1: 1, cos%28pi%2F2%29%2Bred%28i%29%2Asin%28pi%2F2%29+=+red%28i%29,
cos%28pi%29%2Bred%28i%29%2Asin%28pi%29+=+-1 and cos%283%2Api%2F2%29%2Bred%28i%29%2Asin%283%2Api%2F2%29+=+-red%28i%29.
Figure 4 shows complex (real) number 1 in red and three other roots in blue.
These four points are vertices of the square.


Figure 4. 4-th roots
    of the complex number 1



5) 5-th roots of a complex number 1.
In the complex plane 1 has the modulus r=1 and the argument
values alpha=0,2%2Api, 4%2Api, 6%2Api, 8%2Api,....
Taking the 5-th root of the complex number 1, you have the modulus sqrt%281%29=1 (positive value) and                 
five argument values alpha=0, alpha=2%2Api%2F5, alpha=%282%2Api%2B2%2Api%29%2F5=4%2Api%2F5, alpha=%282%2Api%2B4%2Api%29%2F5=6%2Api%2F5 and
alpha=%282%2Api%2B6%2Api%29%2F5=8%2Api%2F5.
This gives five values of the 5-th root of the complex number 1: 1, cos%282%2Api%2F5%29%2Bred%28i%29%2Asin%282%2Api%2F5%29,
cos%284%2Api%2F5%29%2Bred%28i%29%2Asin%284%2Api%2F5%29, cos%286%2Api%2F5%29%2Bred%28i%29%2Asin%286%2Api%2F5%29, and cos%288%2Api%2F5%29%2Bred%28i%29%2Asin%288%2Api%2F5%29.
Figure 5 shows complex (real) number 1 in red and four other roots in blue.
These five points are vertices of the regular pentagon.


Figure 5. 5-th roots
    of the complex number 1



6) 6-th roots of a complex number 1.
In the complex plane 1 has the modulus r=1 and the argument
values alpha=0,2%2Api, 4%2Api, 6%2Api, 8%2Api, 10%2Api,....
Taking the 6-th root of the complex number 1, you have the modulus sqrt%281%29=1 (positive value) and                 
six argument values alpha=0, alpha=2%2Api%2F6=pi%2F3, alpha=%282%2Api%2B2%2Api%29%2F6=2%2Api%2F3, alpha=%282%2Api%2B4%2Api%29%2F6=pi,
alpha=%282%2Api%2B6%2Api%29%2F6=4%2Api%2F3 and alpha=%282%2Api%2B8%2Api%29%2F6=5%2Api%2F3.
This gives six values of the 6-th root of the complex number 1: 1, cos%28pi%2F3%29%2Bred%28i%29%2Asin%28pi%2F3%29,
cos%282%2Api%2F3%29%2Bred%28i%29%2Asin%282%2Api%2F3%29, cos%28pi%29%2Bred%28i%29%2Asin%28pi%29, cos%284%2Api%2F3%29%2Bred%28i%29%2Asin%284%2Api%2F3%29
and cos%285%2Api%2F3%29%2Bred%28i%29%2Asin%285%2Api%2F3%29.
Figure 6 shows complex (real) number 1 in red and five other roots in blue.
These six points are vertices of the regular hexagon.


Figure 6. 6-th roots
    of the complex number 1



Summary
Let w be a complex number r%28cos%28alpha%29+%2B+red%28i%29%2Asin%28alpha%29%29 and n be a positive integer n>=2. Then w has n distinct n-th roots

, k = 0, 1, ...,n-1.

They all have the same modulus root%28n%2Cr%29 (positive real value) and n distinct argument values %28alpha+%2B+2k%2Api%29%2Fn, k = 0, 1, ...,n-1.



For your convenience, below is the list of my relevant lessons on complex numbers in this site in the logical order. They all are under the current topic Complex numbers in the section Algebra II.
Complex numbers and arithmetical operations over them
Complex plane
Addition and subtraction of complex numbers in complex plane
Multiplication and division of complex numbers in complex plane
Raising a complex number to an integer power
How to take a root of a complex number                                                                (this lesson)
Solution of the quadratic equation with real coefficients on complex domain
How to take a square root of a complex number
Solution of the quadratic equation with complex coefficients on complex domain




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