# Question about nth root of unity

1. Apr 22, 2014

### mesa

1. The problem statement, all variables and given/known data
Determine the nth roots of unity by aid of the Argand diagram...

2. Relevant equations

Is the nth root of unity where we have a complex number to the 'nth' power equal to 1? For example,

$$(2+i)^n=1$$

3. The attempt at a solution
None yet, still trying to translate the question.

Last edited: Apr 22, 2014
2. Apr 22, 2014

### SammyS

Staff Emeritus
It's more like:

Solve zn = 1 for z , where z is complex.

3. Apr 22, 2014

### slider142

Yes. For example, $-\frac{1}{2} + i\frac{\sqrt{3}}{2}$ is one of the 3rd roots of unity because $\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right)^3 = 1$.

4. Apr 22, 2014

### SteamKing

Staff Emeritus
In math in English, 'unity' is a quaint way of saying 'one'.

5. Apr 22, 2014

### Staff: Mentor

The only integer n for which this is true is n = 0.

2 + i is a complex number whose magnitude is √5. Multiplying it by itself is effectively rotating it by a certain amount, and increasing the magnitude. In polar form, this is easier to see.
2 + i = √5e, where θ = tan-1(1/2).
(2 + i)n = [√5e]n = (√5)neinθ

6. Apr 22, 2014

### mesa

Okay, this makes much more sense now, so we are looking for x and y values that will have a magnitude (modulus z, or r) of 1 and we can use a generating function for complex numbers like so:

$$\left( \frac{1}{k} \right)+i\left( \frac{\sqrt{k^2-1}}{k} \right)$$

or this,

$$\left( \frac{\sqrt{k^2-1}}{k} \right)+i\left( \frac{1}{k} \right)$$

I worked really hard with Latex on those big parentheses only to decide I prefer them without :)

7. Apr 22, 2014

### Staff: Mentor

This looks really nice - I like the big parens - but I don't see how this is answers your original question. An nth root of unity is a complex number z, such that zn = 1. If you write z = 1e, then you have zn = 1neniθ.

My comment followed your question about whether (2 + i)n could equal 1.

8. Apr 22, 2014

### SammyS

Staff Emeritus
How is k related to n ?

9. Apr 22, 2014

### ChrisVer

the parenthesis are wrong... $z^{n}≠|z|^{n}$
in other words, writing $z^{2}$ you mean $zz$ not $zz^{*}$

So if you take the 2nd root for example (n=2) of the formula you provided:
$(\frac{1}{k}+i \frac{\sqrt{k^{2}-1}}{k})(\frac{1}{k}+i \frac{\sqrt{k^{2}-1}}{k})≠1$...

If you want to use the form for $z=x+iy$
then
$(x+iy)^{n}=1$
can be terrible...but not impossible to show....
Using the Binomial theorem:
$(x+iy)= \sum_{k=0}^{n} (n k) x^{k} (iy)^{n-k}$
http://www.dummies.com/how-to/content/how-to-expand-a-binomial-that-contains-complex-num.html
and proceed accordingly....(probably you'll have to end up with sins and cos, as a taylor expansions?)

Of course that's totally tedious... it's faster to follow the already proposed idea of writing $z=r e^{i\theta} (r=1)$
from the last you can also deduce that $x^{2}+y^{2}=r^{2}=1$
or better put that $(x,y)$ belong on the unitary "circle" (in fact it's a circle for n going to infinity, otherwise it's a "circle" created by n-points). So here again you see the sin,cos solutions (which must be on the complex plane)....

Last edited: Apr 22, 2014
10. Apr 25, 2014

### mesa

Thanks! Maybe they aren't so bad then :)

So I got this from your suggestions,

$$1=\cos \left( \frac{2Pik}{n} \right) + i\sin \left( \frac{2Pik}{n} \right)$$

What a remarkable answer, and so versatile! I used algebra to write another solution but so far have only been able to get one answer for the cubic.

It's not really, went off on a tangent :P

Yup, I messed up and was headed the wrong direction but have it now. I love binomial theorem, I wish they went over it in our engineering program, at least we have breaks for playing with such delightful things.