Where is the mistake in this proof?

Question

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.

Answer 1

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

Answer 2

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."

Answer 3

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