# Complex exponentiation “proof” that $\pi = 0$ [closed]

\begin{align} e^{2 \pi i } &= 1 \\ e^{1 + 2 \pi i} &= e \\ e^{1 - 2 \pi i} &= e \\ {(e^{1 - 2 \pi i})}^{1 + 2 \pi i} &= e^{1 + 2 \pi i}=e\\ e^{1+4 \pi^2} &= e\\ e^{4 \pi^2} &= 1 \\ \pi &= 0 \end{align}

Where is an error?

## closed as off-topic by Namaste, kingW3, Simply Beautiful Art, Lord Shark the Unknown, LeucippusJul 16 '17 at 5:19

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• How did you conclude $e^{4 \pi^2}=1 \implies \pi = 0$? – Sahiba Arora Jul 15 '17 at 19:28
• This strikes me as a more involved version of$$e^{z} = e^{2\pi i(z/2\pi i)} = (e^{2\pi i})^{z/2\pi i} = 1^{z/2\pi i} = 1$$for all complex $z$. – Andrew D. Hwang Jul 15 '17 at 19:33
• @kp9r4d Well that is the mistake. – Sahiba Arora Jul 15 '17 at 19:34
• Exponentiation to complex powers does not follow the multiplication rule, i.e. $(a^b)^c$ need not be equal to $a^{bc}$. – Matt Samuel Jul 15 '17 at 19:37
• No error. Pi does equal zero. It's the sad existential absurdity of the universe. ... Okay, logs of complex numbers are multivalued because exponents are periodic. This result if very similar to $-5 = \sqrt {(-5)^2} = 5$ so $5 = -5$ of that $\arccos (\cos 7\pi/4 ) = pi/4$ so $7 = 1$. $e^{2\pi i} = 3^0 = 1$. That doesn't meant $2\pi i = 0$. Because exponents are not 1 to 1 in reals and therefore logarithms are not unique. – fleablood Jul 15 '17 at 19:38

There are actually three mistakes in the proof:

No 1:

$$e^{1 - 2 \pi i} = e \Rightarrow {(e^{1 - 2 \pi i})}^{1 + 2 \pi i} = e^{1 + 2 \pi i}$$

In the second equation, the LHS is a multi-valued function, while the RHS is a complex number.

No 2:

$$e^{4 \pi^2} = 1 \Rightarrow \pi = 0$$

As complex function, the exponential is NOT one to one.

No 3

Also note that the laws of exponentiation don't hold for complex exponentials, one should not use $${(e^{1 - 2 \pi i})}^{1 + 2 \pi i}= e^{(1 - 2 \pi i) (1 + 2 \pi i)}$$

• But if we additionally knows that $4 \pi^2$ is real then No 2 is not really mistake. – kp9r4d Jul 15 '17 at 19:48
• Am I the only one who hates thinking of complex exponentiation as a "multi-valued" function. It seems really stupid to me and makes it sound unnecessarily complicated and magical. – mathworker21 Jul 15 '17 at 20:29
• I think's so too, it's more reasonable to say that $z \mapsto z^w$ is not really well defined function $\mathbb{C} \to \mathbb{C}$. – kp9r4d Jul 15 '17 at 20:34
• @mathworker21 When you get into stuff like complex analysis, you define complex exponentiation to be single valued (so an actual function), however, you lose common exponential rules that hold for real numbers sometimes. – Simply Beautiful Art Jul 15 '17 at 21:01
• @kp9r4d One usually defines $z\mapsto z^w$ as $z\mapsto e^{w\ln(z)}$, where $\ln(z)$ is usually defined by $\ln|z|+i\operatorname{atan}(z)$, where $\operatorname{atan}(z)\in(-\pi,\pi]$ and $e^{x+iy}=e^x(\cos(y)+i\sin(y))$, a well-defined function. – Simply Beautiful Art Jul 15 '17 at 21:03