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Complex Roots of Unity

September 22nd, 2009

Without complex numbers taking the square root of any positive integer, such as 1, will give you two answers, in this case +/- 1, but taking the cube roots will only five you one answer, 1. However when we consider the complex roots you will find that the nth root of any number will give you n roots.

This can be shown using De Moivres theorem. Firstly consider the complex number z = cosΘ + isinΘ and let

zn = 1
(cosΘ+isinΘ)n = 1

which using De Moivres theorem gives

cos nΘ + isin nΘ = 1

We can now compare the real and imaginary parts to find the values of n

cos nΘ = 1 and sin nΘ = 0

Therefore nΘ = 2pπ for integral values of p
So \theta = \frac{2p\pi}{n}

which gives

\theta = \frac{2\pi}{n} , \frac{4\pi}{n} , \frac{6\pi}{n} , \frac{8\pi}{n} , … , 2\pi

so the roots of unity are
z_1 = cos \frac{2\pi}{n} + i sin \frac{2\pi}{n}
z_2 = cos \frac{4\pi}{n} + i sin \frac{4\pi}{n}
z_3 = cos \frac{6\pi}{n} + i sin \frac{6\pi}{n}
etc

However what is interesting about these is when you plot them on an argand diagram they are evenly spaced around a circle of radius 1 and centred at the origin. An example for n=8 is shown below

The roots of unity are space evenly around a circle of radius 1 and center at the origin

The roots of unity are space evenly around a circle of radius 1 and centre at the origin

This happens because the increase in the angle for each successive root is equal since we divided 2pi by n.

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Comments Left

  1. Peter L. Griffiths
    August 23rd, 2011 at 16:57 | #1
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    It is in fact easier to find the n nth roots of any complex number a+ib. You obtain the arcotangent below 90 degrees of the complex ratio a/b and then divide by the root required. The cotangent of the result gives one of the n roots of the complex ratio. For the other n-1 roots you add 360 degrees to the arcotangent n-1 times. This gives you the n nth roots of the complex number. This system can be reconciled with finding the n nth roots of +1, -1, i and -i by making a in the complex number equal 0 and substituting Cotes’s format cos90+isin90 equals 0+i. It will be found that cos180+isin180 equals -1 and cos360+isin 360 equals i to the power of 4 which is +1. This system also applies for division, cos45+isin45 equals the square root of i.

  2. admin
    August 25th, 2011 at 10:46 | #2
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    @Peter L. Griffiths
    Hi, Thanks for taking the time to comment :)
    The method you outlined is one of the easiest I’ve seem for finding the roots of a complex number

  3. Peter L. Griffiths
    March 11th, 2012 at 18:04 | #3
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    My comments of 23 August 2011 have important implications for Hamilton’s Quaternions equation which is i^2=j^2=k^2=ijk=-1. This equation is incorrect because -1 can only have two square roots, so that k^2 cannot equal -1 unless k is either i or j. Any real, imaginary or complex number can only have two square roots, three cube roots, four fourth roots, five fifth roots etc.