Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$

Obviously, one of my algebraic manipulations is not valid.

share|improve this question
30  
The second equality: $(a^b)^c$ is not equal to $a^{bc}$ when $a,b,c\in\mathbb{C}$. Both expressions are multivalued functions, i.e. they can have several meanings. –  Stefan Hansen Jan 18 '13 at 17:49
3  
@StefanHansen, you can elaborate (not because it is insufficient but to make it worthy of being called answer.) a little and answer it . –  007resu Jan 18 '13 at 18:15
5  
The proof is exactly the same idea as $-1=\sqrt{(-1)^2}=\sqrt{1}=1$. It is less obvious just because the "root" part is the $\frac{1}{2\pi}$ power... –  N. S. Jan 18 '13 at 19:08
4  
Or $1=1^{1/2}=(e^{i 2 \pi })^{1/2}= e^{i \pi} = -1$ –  leonbloy Jan 18 '13 at 19:11

1 Answer 1

up vote 32 down vote accepted

As said in the comments, the expressions $(a^b)^c$ and $a^{bc}$ are in fact multivalued functions; they are not a uniquely determined complex number. A classic example of a multivalued function is the complex logarithm denoted by $\log(z)$, $z\in\mathbb{C}$. The complex logarithm $\log(z)$ is any complex number $w$ satisfying $e^w=z$ (which has several solutions, see e.g. this), and hence $\log(z)$ gives rise to a whole set of complex numbers instead of just a single complex number.

Complex exponentiation such as $z^w$ for $z,w\in\mathbb{C}$ is usually defined as $$ z^w=\exp(w\log(z)), $$ where $\log(z)$ is the complex logarithm, and hence this is also a multivalued function. I hope this sheds some light on the problems with doing manipulations on complex numbers as if they were real numbers. See also this for other examples of identities which fail when using complex numbers as they were real numbers.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.