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It is standard practice to define on $\mathbb{C}$, $$\operatorname{Log}(z) = \log(|z|) + i \operatorname{Arg}(z).$$ When composed with $\exp$, we get $\operatorname{Log} \circ \exp (z) = z$, the identity function, for all $z$ in the $2\pi $-wide strip $\{ z\, :\, 0 < \Im(z) < 2\pi \}$.

Now, on the one hand, in case two analytic functions are identical on an open set, then they are identical. On the other hand, $\operatorname{Log} \circ \exp$ is certainly not the identity function throughout its domain. Where is the faulty deduction ?

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up vote 2 down vote accepted

You get $\;\; (\operatorname{Log} \circ \exp)(z) \: = \: z \;\;$ for all $z$ in the $2\pi$-wide strip $\: \{z : -\pi < \operatorname{Im}(z) < \pi\} \:\:$.

The faulty deduction would be any conclusion that $\: \operatorname{Log} \circ \exp \:$ is an analytic function.
$\operatorname{Log} \circ \exp \:$ is either not continuous or not even a function, depending
on whether you define $\operatorname{Arg}$ to be a function or multi-valued.

(Also, the analytic functions $\: z\mapsto 1 \:$ and $\: z\mapsto 0 \:$ are identical on the empty set, which is open.)

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Thanks Ricky. Regarding the empty set, you're right of course. Should have said "a non-empty open set". Regarding $Arg$, you mean it's not continuous on (say) the negative real axis. Still, $Log$ will be analytic on $\mathbb{C} \setminus \{x + 0 i \, : \, x \le 0 \}$. – Teddy Jul 16 '12 at 11:20

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