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Given

$$e^{e^x}=1$$

Wolfram gave me the answer

$$x=\ln{(2i\pi n_1)}+2i\pi n_2\quad n_1,\,n_2 \in \Bbb N$$

doesn't that mean if i have $e^x=1$, The answer is $x=\ln(1)+2i\pi n, n\in \Bbb N$.

But honestly, it bothers me. Why we add an extra $2i\pi n$ after we take the logarithm? Is it complex logarithm or what? Cz, yes, i know $e^x=0$ has no solutions in real plane. But i have a feeling it's a complex logarithm.

Please explain to me with easy language. Thanks in advance.

user516076
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    Yes it's because$$e^{z+2i\pi n}=e^z\qquad\forall n\in\mathbb{Z}$$ – Peter Foreman May 22 '20 at 08:55
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    To begin with, you need the definition of complex exponential by Euler's formula: https://en.wikipedia.org/wiki/Euler%27s_formula. Then try to solve $e^x=0$. – Miguel May 22 '20 at 08:57

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The complex logarithm is a multivalued function. This is the reason of your inifnitely many solutions. More precisely, given $z=\rho e^ {i \theta} \in \mathbb{C} $, we have: $$ \ln z:= \ln \rho + i (\theta + 2 n \pi) $$ with $n \in \mathbb{N}$. If, for instance, $z=1$, the complex logarithms are: $$ \ln 1 = 0 +2n \pi i = 2 n \pi i $$

Manuel Norman
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