# Is i = 0 from Euler's Identity? [duplicate]

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# ⇒ $$i = 0$$

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

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

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• Logarithms and powers of complex numbers cannot be handled the same you handle logarithms and powers of real numbers. – Kabo Murphy Oct 18 at 8:38
• 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
• 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. – peawormsworth Oct 18 at 9:13
• It is true that i has 0 real part. – peawormsworth Oct 18 at 9:19

## 1 Answer

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.