# Euler's identity

1. Oct 14, 2009

### SW VandeCarr

$$e^{i\pi}=-1$$

$$e^{i\frac{\pi}{2}}=i$$

but

$$e^{i\frac{\pi}{3}}\neq-1$$

I know there are infinitely many solutions here, but I would expect the third result should include -1 as the cube root of itself. However $$e^{\pm ix}=cos(x)\pm{isin(x)}$$ would not seem to give -1 for any solution for $$x=\frac{\pi}{3}$$. Where am I going wrong here?

2. Oct 14, 2009

### zpconn

I'm not sure what you mean by "infinitely many solutions."

If $$z = r\exp(i\theta)$$ with $$r > 0$$ and $$\theta \in [0,2\pi)$$, then $$z = w^n$$ has exactly $$n$$ solutions given by

$$r^{1/n}\exp(i(2\pi k + \theta)/n)$$

for $$k=0,1,\dots,n-1$$. You can check that $$\exp(i(\pi/3))$$ is one of these for $$z = -1$$ and $$n=3$$, so you have found a cube root of -1, but there are two others, one of which is -1. Notice the above formula shows that each of the three cube roots of -1 have different angles (and they are equally spaced on the unit circle in the complex plane).

3. Oct 15, 2009

### SW VandeCarr

Thanks zpconn for your informative response. However, I wouldn't expect that just defining the problem in terms of polar coordinates could change the fundamental nature of the problem. It should be understandable in terms of Cartesian coordinates, or am I wrong here?

Algebraically if we have $$exp(\pm{i\pi})=-1$$, then taking the cube root of both sides should yield -1, not just a complex number. To get -1 as a solution we must have $$cos(x)=-1$$ and $$isin(x)=0$$. This true for $$x=\pm(\pi,3\pi,5\pi......)$$ for all odd multiples of $$\pi$$. This is what I meant by an infinite number of solutions.

Last edited: Oct 15, 2009
4. Oct 15, 2009

No the number of solutions is not infinite, because all of those solutions that you show are the same complex number. You abuse the algebraic notation and didn't go through the work of defining a good cube root function on the complex numbers.

If you want something more algebraical: Every polynomial of degree n can be factored by it's complex roots. You cannot factor:
$$x^3+1=0$$
with (x+1) alone. You need three roots, the other two are:
$$\frac{1}{2}\pm i\frac{\sqrt{3}}{2}$$
One of these solutions is what you have discovered. It is also a cube root of -1. Just one you don't like. Why did you expect -1 to be the only number that produces -1 when you multiply it three times with itself?

5. Oct 15, 2009

### mathman

There are three cube roots. -1 is one of them, the other two are complex conjugates
given by $$e^{i\frac{\pi}{3}}$$ and $$e^{-i\frac{\pi}{3}}$$.

6. Oct 15, 2009

### SW VandeCarr

Yes, I understood there were two complex roots gotten by the substitution of $$x=\pm\frac{\pi}{3}$$ into the Euler equation. However this equation only gives the real root -1 when x is equal to odd integer multiples of $$\pi$$. I don't understand why this is the case. Is this equation insufficient under certain circumstances such as not defining the problem in terms of polar coordinates?

Last edited: Oct 15, 2009
7. Oct 15, 2009

### zpconn

The real issue here I suspect is that as a function the cube root should select just one cube root, but there are three, so without a proper understanding the situation can be very confusing (as it is here). This is a well-known and well-studied phenomenon in complex analysis, and there are several ways of understanding it. The best is through Riemann surfaces, but that's of course a bit much and not strictly necessary.

The cube root is a special case of complex exponentiation, and exponentiation is defined in terms of the complex exponential and the complex logarithm. So I'll take the exponential function as already defined.

The polar perspective really is a convenient way to look at this. Everything gets interesting when we try to define the complex logarithm because angles can be written in infinitely many different ways. The complex logarithm has infinitely many branches; the $$n$$th branch is

$$\log(r\exp(i\theta)) = \ln(r) + i \theta + 2\pi i n$$

where we impose $$\theta \in [0,2\pi)$$.

This follows from Euler's formula. The branch $$n=0$$ is the principal branch $$\ln(r) + i\theta$$. So complex exponentiation becomes branched as well because we define for arbitrary $$z,k \in C$$ the exponentiation operation

$$z^k = \exp(k\log(z))$$,

inspired by the analogous relation for real numbers. Now, if you write $$z = r\exp(i\theta)$$ with $$\theta \in [0,2\pi)$$ as before and $$k=a+bi$$, some algebra will show the $$n$$th branch of the complex exponentiation is given by

$$\exp(2n\pi i a) \exp(-2n\pi b) [z^k]$$

where $$[z^k]$$ is just meant to represent the principal branch of the exponentiation, i.e., this gives the $$n$$th branch in terms of the principal branch (assuming I didn't make mistakes).

I've written the result in this form so that it's easy to see that if $$b$$ is nonzero, then the term $$\exp(-2n\pi b)$$ gives rise to infinitely many branches, i.e., the exponentiation will not return to its principal branch as $$n$$ runs over the positive integers.

In your case, $$b = 0$$ and $$a=1/3$$. The term $$\exp(2n\pi i a)$$ shows you that there will be precisely three branches.

I am out of time, so I can't finish this, but hopefully I've helped elucidate this mystery for you. It's not hard to finish from here.

8. Oct 15, 2009

### SW VandeCarr

Many thanks zpconn. I appreciate you taking the time to write this all out. I knew their was more to the problem then just substituting values for x in cos(x) and (i)sin(x) in Euler's equation, but didn't know exactly how to go about it. It seems that how the problem is defined (polar coordinates) is crucial to getting all the roots. One question: Since we are dealing with the unit circle, can I assume r=1?

Last edited: Oct 15, 2009