Exercise [05.10]

ZZR Puig · 17 Jun 2008, 12:12

As a result of the discussion held on the exercise discussion board about this paradox I'm completely rewriting my previously posted solution.

Preface

Before facing the solution of the paradox itself I think it may be worth to make a brief clarification about the exercise wording or presentation. That is:

$e=e^{1+i2\pi}$ therefore $e=(e^{1+i2\pi})^{1+i2\pi}=e^{1+i4\pi-4\pi^2}=e^{1-4\pi^2}$

As I stated in my previous paradox solution proposal, I think that this formulation can be misunderstood. In the way that the first equation don't necessarily involve the first equality of the second expression.

The initial equation can be understood as $e=ee^{i2\pi}=ee^{i2\pi}e^{i2\pi}...=e^{1+ki2\pi} \Rightarrow e=e^{1+ki2\pi}$

While the second one could be generalized as $e=(e^{1+i2\pi})^{1+i2\pi}=e^{(1+i2\pi)^k} \Rightarrow e=e^{(1+i2\pi)^k}$

I think it's quite obvious that $e^{1+i2k\pi}\neq{e^{(1+i2\pi)^k}$ even if the are equal for k=1

.

Main solution

Nonetheless the paradox continues to exist if we consider that ${\left(e^{1+i2\pi}\Right)^{1+i2\pi}$ can have two different solutions depending on whether we solve the inner parenthesis first or we multiply the exponents.
$a) \qquad ({e})^{1+i2\pi}=e$
$b) \qquad {e^{1-4\pi^2+i4\pi}}=e^{1-4\pi^2}$

Let's resolve this paradox then.

We know that $w^z=e^{w+ik2\pi}$ or that $\log {(w^z)}=(z+ik2\pi) \cdot \log {w}$ . So let's apply logarithms to the original expression.

$e=(e^{1+i2\pi})^{1+i2\pi}$
$\log {e}=\log {\left((e^{1+i2\pi})^{1+i2\pi}\right)}$

Then,
$1+ik2\pi=(1+i2\pi+iC2\pi)\cdot \log {(e^{1+i2\pi})}$
$1+ik2\pi=(1+i2\pi+iC2\pi)\cdot (1+i2\pi+iD2\pi)$ , where

,

and

are integers.

For simplicity we redefine A=C+1