# If $e^{2\pi i}=1$ then is $2\pi i=0$?

So with Euler identity I used $$2 \pi(\tau)$$ instead and got $$e^{2\pi i} = 1$$, and took natural logarithm $$\ I 2\pi = 0$$? But is see online the answer is $$6.28....i$$.

• The logarithm of a complex number is not unique, but has an infinite number of values separated by $\ 2 \pi \ i \$ . So zero is just one possible value. You can see this in what you wrote: $\ e^0 \$ also equals 1 , and so does $\ e^{4 \pi i } \ , \ e^{-2 \pi i } \$ , etc. – colormegone Feb 21 '16 at 3:39
• en.wikipedia.org/wiki/… – pjs36 Feb 21 '16 at 3:39
• Interesting I'll have a read thanks – jake walsh Feb 21 '16 at 3:40
• $(-1)^2 = 1^2$ so $-1 = 1$? Obviously not. Similarly, what you wrote isn't true also, it just means when you allow for complex number solutions, $e^x=1$ has solutions other than $x=0$. – Deepak Jul 19 at 12:54

In $$(0,\infty)$$, every number has one and only one real logarithm. So, in $$(0,\infty)$$, it makes sense to assert that $$e^x=y\iff x=\log y$$.
However, every complex numbers has infinitely many logarithms. In particular every number of the form $$2k\pi i$$ (with $$k\in\mathbb Z$$) is a logarithm of $$1$$. So, your approach does not work here.
Another way of seeing this is: in $$\mathbb R$$, the exponential function is injective, but not in $$\mathbb C$$.