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This question already has an answer here:

From Euler's identity

$e^{i\pi} = -1$

$e^{2\pi i} = 1$

$[e^{2\pi}]^i = 1$

$i = \log_ {e^{2 π }}1$

$i = 0$

But how is this possible? Please help me in finding that where i am going wrong.

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marked as duplicate by Hans Lundmark, Community Oct 18 at 10:55

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    $\begingroup$ Logarithms and powers of complex numbers cannot be handled the same you handle logarithms and powers of real numbers. $\endgroup$ – Kabo Murphy Oct 18 at 8:38
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    $\begingroup$ The exponential function is not injective on the complex numbers, so the inverse function does not straightforwardly exist (though there are ways of defining a single-valued inverse). You wouldn't say $1=-1$ because $1^2=(-1)^2$ (though there are "paradoxes" built on just that) $\endgroup$ – Mark Bennet Oct 18 at 8:45
  • $\begingroup$ log e^2pi 1 = ln 1 / ln e^2pi ... ln 1 = Real ... e^2pi = Real ... ln e^2pi = Real ... ln 1 / ln e^2pi = Real / Real = Real ... The provided log equation is false, since log real real is real. $\endgroup$ – peawormsworth Oct 18 at 9:13
  • $\begingroup$ It is true that i has 0 real part. $\endgroup$ – peawormsworth Oct 18 at 9:19
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A function $f:A\to B$ is said to be injective if $f(a_1)=f(a_2)$ implies $a_1=a_2$ for $a_1,a_2\in A$.

In other words, an injective function is such that $a_1\neq a_2$ implies $f(a_1)\neq f(a_2)$, i.e. different elements of the domain are mapped to different elements in the codomain. However, plenty of functions are not injective, such as $f(x)=x^2$ when defined over the real numbers: since $(-1)^2=1^2,$ but $-1\neq 1$.

Your argument is essentially $$e^{2\pi i}=1=e^0$$ hence $2\pi i=0$ and $i=0$. The fallacy is in assuming that $e^{2\pi i}=e^0$ implies $2\pi i=0$. Think of $f(z)=e^z$ as a function. What you are saying is that $f(2\pi i)=f(0)$, so $2\pi i=0$, which might not be true in general! Indeed, it is not true here as the exponential function is not injective.

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