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Here's a "proof" of $e^x=1$ for all $x$: $$\exp(x)=\exp\left(i2π⋅\frac{x}{i2π}\right)=\bigl(\exp(i2π)\bigr)^{x/(i2π)}=1^{x/(i2π)}=1$$ Why is the second equality wrong?

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Don't post again the same question: edit the former one...and use LaTeX! –  DonAntonio Apr 11 '13 at 18:42
    
possible duplicate of Can someone tell me why the second equality is wrong? –  DonAntonio Apr 11 '13 at 18:43
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The "double power" rule for exponents does not hold true for complex expressions, just like there are more rules from real algebra that do not "copy" to algebra involving complex numbers –  imranfat Apr 11 '13 at 18:45
    
@DonAntonio this one makes slightly more sense –  Lost1 Apr 11 '13 at 18:45
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To best understand the explanation (and to get the best answers), it would be useful to first explain why you think the second equality is right! –  Hurkyl Apr 11 '13 at 18:50

1 Answer 1

This argument is false because $h(u,v)=u^v$ is not a single-valued function in the complex numbers. In particular, $1^v\neq 1$ for some values of $1^v$.

If you choose some single-valued branch of $u^v$, then it is not true that $\exp(xv)=\exp(x)^v$ in general.

The wonderful, magical thing about $\exp(x)$ is that it is a function - it is well-defined for all complex $x$. But in general exponentiation, we are not so lucky. If $a$ is a complex number, then we usually define $a^x$ to be $\exp(x\ln a)$, but $\ln a$ is not well-defined - there are infinitely many different possible values for $\ln a$.

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Thank you for explaining how ln(a) is not well defined for complex number. –  lakshman.pasala Apr 11 '13 at 20:23

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