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 $6.28....i$.

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    $\begingroup$ 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. $\endgroup$ – colormegone Feb 21 '16 at 3:39
  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – pjs36 Feb 21 '16 at 3:39
  • $\begingroup$ Interesting I'll have a read thanks $\endgroup$ – jake walsh Feb 21 '16 at 3:40
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    $\begingroup$ $(-1)^2 = 1^2$ 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 $x=0$. $\endgroup$ – 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 $\mathbb C$.


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