# ⇒ $$i = 0$$

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

• 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) – Mark Bennet Oct 18 at 8:45
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.