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

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2 Answers

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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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The exponential function $f(x)=e^x$ is a group homomorphism from $(\mathbb{C},+)$ to $(\mathbb{C} \setminus \{0\}, \cdot)$. And yes, this group homomorphism is not injective, for $f(2\pi{i})=f(0)=1$. In fact, $\operatorname{Ker}(f)=2\pi{i}\mathbb{Z}$.

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