# Thread: e to the i pi misunderstanding

1. ## e to the i pi misunderstanding

Dunno if this should go in this subforum, but I came across it working on a differential equation.

So e^(xi) = cosx + isinx,
therefore e^(i*pi) = -1.
And (-1)^(1/(1+\-2k)) = -1, where k is any integer.

Now, z^(a*b) = z^(a^(b)) = z^(b^(a)),
so
e^(i*pi*(1/3)) = e^((i*pi)^(1/3))

Since e^(i*pi) = -1,
e^((i*pi)^(1/3)) = (-1)^(1/3) = -1

But that isn't correct, I know, because the worked-out solution to this problem from cramster.com, my calculator, and the answer in the back of my textbook all say to me that e^((i*pi)^(1/3)) = 1/2 + (sqr(3)/2)i.

My calculator even knows that this -1 obtained from raising e to the i pi isn't the same as a -1 obtained from simply punching it in. I put e^(i*pi) on my calculator, got -1 as an answer, and stored it as A. Then I put in -1, got -1 as an answer, and stored it as B. Then I put in A^(1/3) and B^(1/3), and got 1/2 + (sqr(3)/2)i and -1, respectively.

(Also, I raised e to pi, then raised that answer to i, and got -1 + 1E-13i, really close to -1, but not actually it. I know that calculators - I have a TI-84 Plus - might do things more arithmetically than a human, just to be simple, and so might obtain slightly different answers than one would by using exact theory, but still...)

So... e^(i*pi) doesn't REALLY equal -1? What's going on here? What is the reasoning behind why e^(i*pi*(1/3)) = 1/2 + (sqr(3)/2)i (or possibly 1/2 +\- (sqr(3)/2)i)?

2. Hello,

(-1)^(1/3) = -1
This equality is false...

let x be the possible values for $(-1)^{1/3}$

we have $(-1)^{1/3}=x$
so $x^3+1=0$

this factors into $(x+1)(x^2-x+1)=0$
so $x=-1$ or $x^2-x+1=\left(x-\frac{1-i \sqrt{3}}{2}\right)\left(x-\frac{1+i \sqrt{3}}{2}\right) \Rightarrow x=\frac{1-i \sqrt{3}}{2} \text{ or } x=\frac{1+i \sqrt{3}}{2}$

more generally, the third roots of the unity, $e^{i \pi/3}~,~ e^{2i \pi/3} ~,~ e^{i \pi}$ satisfy the equation $x^3=-1$

you could have written :
$e^{i\pi/3}=\cos \left(\tfrac \pi 3\right)+i \sin \left(\tfrac \pi 3\right)=\frac 12+i \frac{\sqrt{3}}{2}$

3. That makes sense.

That x has three roots if x = (-1)^(1/3) is one of the things I forgot here.

Thanks, that was a simple and concise explanation.