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Homomorphism from complex numbers with addition to the non zero complex numbers with multiplication injective? Surjective?

I have been given the map $\phi:\mathbb{C}\rightarrow \mathbb{C}-\{0\}$ by $\phi:z \mapsto e^z$.

I have shown the map is not injective.

I'm not sure about surjectivity though?

How do I prove it is surjective (or not)?

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    $\begingroup$ Why $e^z =1 \iff z= 0$? That's the definition of a trivial kernel, but you haven't shown it. (in fact it's not true.) $\endgroup$ May 14, 2017 at 20:19
  • $\begingroup$ What about $e^{2\pi i}$? $\endgroup$
    – sharding4
    May 14, 2017 at 20:19
  • $\begingroup$ Oh okay my bad. So $e^{2\pi i}=1$ therefore the kernel is not trivial and so the map cannot be injective. What about surjectivity? $\endgroup$
    – Ben B
    May 14, 2017 at 20:21
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    $\begingroup$ If it's not injective it's not an isomorphism. You don't also need to find out whether or not it is surjective (unless you are specifically asked to, or you are just curious.) $\endgroup$ May 14, 2017 at 20:23
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    $\begingroup$ Note that the fact that this specific map is not an isomorphism does not prove that the sets are not isomorphic. You would need to show that there is no possible isomorphism. $\endgroup$ May 14, 2017 at 20:24

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