# Complex Number Graph : x^5 = 1

1. Jun 23, 2011

### jlpmghrs

Well this is the question...and I don't understand it very well :

Suppose you know that x raised to the fifth power is 1. What can you say about x's absolute value? (remember that the absolute value of a product is the product of the absolute values of its factors.) What can you say about x's angle? (remember that the angle of a product is the sum of the angles of its factors.)

This is what I based my first graph on :
the 1st statement means : x5 = 1 ....right?
2nd : well I can say it's absolute value is 1 since the power is odd, and therefore, |x| = 1. But I didn't do any product of factors (what factors?!?! ...do they mean 1*1*1*1*1?). I mean I don't get what they're asking AT ALL!!
3rd : well the graph shows a zero degree angle. and we can also get 0 by doing
(imaginary number part/real number part) = (0/1) = 0. but again here, what factors are they talking about?

Now, this is about complex numbers right?? So that's what made me do the second graph, although i don't see it as correct at all, since it would be (x+i)5. Well, I basically did some rubbish here. I'm not sure what even made me do it.

This has really been killing me. I know it's probably really really simple. But I feel like such a fool for not even understanding a bit of the question

Last edited by a moderator: Apr 26, 2017
2. Jun 23, 2011

### micromass

You seem to think that 1 is the only number such that x5=1. This is not true. For example

$$\cos(\frac{2\pi}{5})+i\sin(\frac{2\pi}{5})$$

also has this property. In fact, there are 5 such a numbers such that x5=1.

Let me solve the first problem for you to let you see what they're after. If we know that x5=1 and if we take the absolute value of these things then we get

$$|x^5|=|x|^5=1$$

But now we are working with real numbers, and there is only one real number such that it yields 1 when raised to the fifth power. Thus |x|=1.

Now, can you do something similar with the angles?

Last edited by a moderator: Apr 26, 2017
3. Jun 23, 2011

### I like Serena

Welcome to PF, jlpmghrs!

Yes. This is about complex numbers.
It seems as if you're lacking a bit of knowledge about how those work.
To learn that, you really need a book or perhaps the wiki page.

I'll give you a few pointers though to help with your immediate problem.

Any complex number z can be represented in the X-Y plane as a point, with a distance r to the origin, and with an angle phi with the positive x-axis (the "real" axis).

We write it as:
$$z = r e^{i \phi}$$

Yes, that is the "regular" exponential function, and calculations with it work exactly like normal calculations with the exponential function.
(For now, I'm assuming you know about exponentiation and the mathematical constant e. )

In particular the absolute value of a complex number is the distance to the origin, which is r.
Obviously this is always at least zero.

In your case you have $z^5 = 1$,
meaning:
$$z^5 = (r e^{i \phi})^5 = r^5 e^{5 i \phi}$$

This means that $z^5 = 1$ has a total of 5 solutions of which only 1 is real.
The solutions are:
$$1, \quad e^{\frac {2\pi i} 5}, \quad e^{\frac {4\pi i} 5}, \quad e^{\frac {6\pi i} 5}, \quad e^{\frac {8\pi i} 5}$$

In particular the absolute value is 1.

4. Jun 24, 2011

### gsal

"I like Serena": why is it a total of 5 solutions? Why can't you keep going like:
$$\quad e^{\frac {10\pi i} 5}, \quad e^{\frac {12\pi i} 5} ...$$

5. Jun 24, 2011

### micromass

Because they don't give new solutions. For example: $e^{\frac{10\pi i}{5}}=1$. Remember that $\frac{10\pi}{5}$ denoted an angle. And it turns out that that angle is equal to 0:

$$\frac{10\pi}{5}=2\pi=0$$

Thus $e^{\frac{10\pi i}{5}}=e^0=1$.

The other thingies will also not give new solutions...

6. Jun 24, 2011

### jlpmghrs

Hey thank you guys soo much :) I'm new here and I'm really happy to have help :D

@gsal : according to demoivre's theorem the nth roots of unity are given by : cis (2kpi/n), where k = 0, 1, 2,...., (n-1)

So here,

z5 {n being 5} = Cos(0) + iSin(0), Cos(1.2.pi/5) + iSin(1.2.pi/5), Cos(2.2.pi/5 ) + iSin(4pi/5), Cos(3.2.pi/5) + iSin(6pi/5), Cos(4.2.pi/5) + iSin(8pi/5)

= 1, cis(2pi/5), cis(4pi/5), cis(6pi/5), cis(8pi/5)

And why it only goes upto (n-1) is explained by micromass.

i.e. Cos(5.2.pi/5) + iSin(5.2.pi/5) = Cos 2pi + iSin 2pi = (which gives you again) 1.

Thanks again to everyone