Where is the mistake in this proof?

I can't figure out, where is the mistake:

$$z=re^{i\phi}=re^{\large \frac{2\pi i\phi}{2\pi}}=r(e^{2\pi i})^{\large\frac{\phi}{2\pi}}=r1^{\large\frac{\phi}{2\pi}}=r1=r$$

And we found that the complex numbers are actually real, that can't be true.

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en.wikipedia.org/wiki/… – lab bhattacharjee Jan 30 '13 at 14:52

In the complex domain it is not generally true that $(e^z)^w= e^{zw}$. It is only true if $w$ is element of $\mathbb Z$.

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The "normal rules" of exponents that you are using do not necessarily apply in the complex domain, in particular, when $i \in \mathbb{C}$ is involved.

For $z \in \mathbb{C}\setminus \mathbb{R},\; e^{zu} \neq (e^z)^u$ when $u \notin \mathbb{Z}$.

Specifically, in your case, $\;z = re^{\Large \frac{2\pi i\phi}{2\pi}} \not\rightarrow r(e^{2\pi i})^{\Large\frac{\phi}{2\pi}}$

See in particular this link: Wikipedia: "Some identities for powers and logarithms for positive real numbers will fail for complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions."

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Xpast: I'd suggest taking a quick look at the link to Wikipedia. It explains exactly what does and doesn't carry over to complex numbers with imaginary numbers in exponents, as well as the impact this has on the rules for logarithms (and why this is the case). It may save some headaches in the future! – amWhy Jan 30 '13 at 15:12

$$z=r\exp(i\phi)=r\exp\left(\frac{2\pi i\phi}{2\pi}\right)\stackrel?=r\exp(2\pi i)^{\frac\phi{2\pi}}$$

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