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Conjugate Roots of Complex Numbers

Date: 12/01/2000 at 21:22:23
From: Dennis A. Bailey
Subject: Conjugate roots of complex numbers

Hello. This is my first time using Dr. Math so please bear with me as 
I try and explain the question posed to my class. 

I have taken Algebra I and II and Geometry and I am currently in an 
advanced precalculus course. We have just finished a chapter that 
deals with polar coordinates and complex numbers. My teacher wants to 
know: if you take the nth root of a complex number (a+bi), is there a 
way to tell if there will be any conjugate roots present in your n 

We learned that the nth roots of a complex number when graphed on an 
Argand diagram will produce a regular polygon with n sides. We know 
DeMoivre's theorem and that the n nth roots of z = r*cis(theta) are: 

   z^(1/n) = r^(1/n)*cis[(Theta/n)+(k(360)/n)],  k = 0, 1, ..., n-1

So if you have z^(1/n), z = (a+bi), can you tell whether you will get 
any conjugate roots in the n answers before solving?

My teacher has permitted me to ask you because he is interested in 
your response. He says that he has never seen this question posed in 
any book and that in his seven years of teaching precalculus only one 
person has gotten the answer. Any help that you can give would be 
greatly appreciated.  

sqrt(-1) love math!

Date: 12/01/2000 at 23:41:30
From: Doctor Peterson
Subject: Re: Conjugate roots of complex numbers

Hi, Dennis.

My first thought is to use symmetry. Picture the n nth roots of some 
complex number z; they lie equally spaced around a circle centered at 
the origin. If two of them are conjugates, then the x-axis will be a 
line of symmetry for this set of points; not only that pair, but every 
root whose conjugate is among the roots.

Now I can picture two cases, depending on whether n is even or odd. If 
n is odd, the roots must look like this:

            |                   |
            | *               * |
       *    |                   |    *
            |                   |
    --------+-----*--   --*-----+--------
            |                   |
       *    |                   |    *
            | *               * |
            |                   |

One of the roots will be a positive or negative real (since one of the 
roots must be its own conjugate, in order to have an odd number of 
roots). What numbers have a real nth root? Real numbers. (If you raise 
any of the roots in my pictures to the nth power, multiplying its 
angle by 5, you'll end up on the real axis - positive if there is a 
positive root, negative if there is a negative root.)

How about if n is even? Now all the roots can pair off, and we don't 
have to have a real root, though we might:

            |                   |
         *  |  *              * | *
            |                   |
            |             *     |     *
    --*-----+-----*--   --------+--------
            |             *     |     *
            |                   |
         *  |  *              * | *
            |                   |

Again, raise any root to the nth power, and you see that we end up on 
the positive x-axis for the first case, and on the negative x-axis for 
the second case, just as before.

So the answer is, only real numbers have conjugate pairs of roots. I 
haven't proved anything yet, just thought visually and used examples 
which I can tell will cover all the cases; but I've convinced myself. 
That's how a lot of math is done - first see it, then prove it. (Also, 
I didn't really have to go through all this, but felt like playing a 
bit to see how things work. A lot of math is play.)

So how can we prove this? I've seen that everything comes down to the 
angles involved; so let's move the problem from the complex domain 
into angles. Two roots are conjugate if their angles are opposite. A 
number is an nth root of z if (its length is right, and) n times its 
angle is the angle of z (plus or minus a multiple of 2 pi, of course). 
So if the angle of one root is t, then both nt and -nt must be 
equivalent to the angle of z, and

     nt = -nt + 2 k pi  for some integer k

     2nt = 2 k pi

     nt = k pi

Since this is the angle of z, z must be on the positive or negative 
real axis, as I said.

Putting this back into pictures, we can just draw two conjugate 
numbers and raise them both to the nth power, rotating them in 
opposite directions. If they are both roots of the same number, then 
those multiplications of the two angles must meet somewhere; but the 
symmetry of the situation says that they can only meet on the x-axis. 
Pretty simple, isn't it?

So the real work took a lot less than the play, but wasn't as much 

- Doctor Peterson, The Math Forum
Associated Topics:
College Imaginary/Complex Numbers
High School Imaginary/Complex Numbers

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