It follows from Argument of Product equals Sum of Arguments that the $\arg \left({z}\right)$ function for all $z \in \C$ satisfies the relationship: :$\arg \left({z_1 z_2}\right) = \arg \left({z_1}\right) + \arg \left({z_2}\right)$ which means that $\arg \left({z}\right)$ is a kind of logarithm, in the sense that it satisfies the fundamental property of logarithms: :$\log x y = \log x + \log y$ Notice that $\arg \left({z}\right)$ can not be considered a generalization to complex values of the ordinary $\log$ function for real values, since for $x \in \R$, we have: :$0 = \arg \left({x}\right) \ne \log x$ If we do wish to generalize the $\log$ function to complex values, we can use $\arg \left({z}\right)$ to define a set of functions: :$\operatorname{alog} \left({z}\right) = a \arg \left({z}\right) + \log \left\vert{z}\right\vert$ for any $a \in \C$, where $\left\vert{z}\right\vert$ is the modulus of $z$. All functions satisfy the fundamental property of logarithms and also coincide with the $\log$ function for all real values. This is established in the following lemma. === Lemma 1 === For all $a,z \in \C$, define the (complex valued) function $\operatorname{alog}$ as: : $\operatorname{alog} \left({z}\right) = a \arg \left({z}\right) + \log \left\vert{z}\right\vert$ then, for any $z_1, z_2 \in \C$ and $x \in \R$: : $\operatorname{alog} \left({z_1 z_2}\right) = \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_2}\right)$ and: : $\operatorname{alog} \left({x}\right) = \log x$ This means that our (complex valued) $\operatorname{alog}$ functions can genuinely be considered generalizations of the (real valued) $\log$ function. === Proof of Lemma 1 === Let $z_1, z_2$ be any two complex numbers, straightforward substitution on the definition of $\operatorname{alog}$ yields: {{begin-eqn}} {{eqn | l = \operatorname{alog} \left({z_1 z_2}\right)       | r = a \arg \left({z_1 z_2}\right) + \log \left\vert {z_1 z_2} \right\vert }} {{eqn | r = a \left({\arg \left({z_1}\right) + \arg \left({z_1}\right)}\right) + \log \left({\left\vert {z_1} \right\vert \left\vert {z_2} \right\vert}\right) }} {{eqn | r = a \left({\arg \left({z_1}\right) + \arg \left({z_1}\right)}\right) + \log \left\vert {z_1} \right\vert + \log \left\vert {z_2} \right\vert }} {{eqn | r = a \arg \left({z_1}\right) + \log \left\vert {z_1} \right\vert + a \arg \left({z_2}\right) + \log \left\vert {z_2} \right\vert }} {{eqn | r = \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_1}\right) }} {{end-eqn}} Second part of our lemma is even more straightforward since for $x \in \R$, we have: :$\arg \left({x}\right) = 0$ Then: {{begin-eqn}} {{eqn | l = \operatorname{alog} \left({x}\right)       | r = a \arg \left({x}\right) + \log \left\vert{x}\right\vert }} {{eqn | r = \log x }} {{end-eqn}} which concludes the proof of Lemma 1. {{qed|lemma}} We're left  with an infinitude of possible generalizations of the $\log$ function, namely one for each choice of $a$ in our definition of $\operatorname{alog}$. The following lemma proves that there's a value for $a$ that guarantees our definition of $\operatorname{alog}$ satisfies the much desirable property of $\log$: :$\dfrac{\mathrm d \log x} {\mathrm d x} = \dfrac 1 x$ === Lemma 2 === Let $\operatorname{alog} \left({z}\right) = a \arg \left({z}\right) + \log \left|{z}\right|$. Then if: : $\dfrac {\mathrm d \left({\operatorname{alog} z}\right)} {\mathrm d z} = \dfrac 1 z$ we must have: : $a = i$ === Proof of Lemma 2 === Let $z \in \C$ be such that: : $\left\vert{z}\right\vert = 1$ and: : $\arg \left({z}\right) = \theta$ Then: : $z = \cos \theta + i \sin \theta$ Plugging those values in our definition of $\operatorname{alog}$: {{begin-eqn}} {{eqn | l = \operatorname{alog} \left({z}\right)       | r = a \arg \left({\cos \theta + i \sin \theta}\right) + \log \left\vert{z}\right\vert }} {{eqn | r = a \theta + \log 1 = a \theta }}   {{end-eqn}} We now have: :$a \theta = \operatorname{alog} \left({\cos \theta + i \sin \theta}\right)$ Taking the derivative with respect to $\theta$ on both sides, we have {{begin-eqn}} {{eqn | l = \frac{\mathrm d}{\mathrm d \theta} (a \theta)       | r = \frac{\mathrm d}{\mathrm d \theta} \left({\operatorname{alog} \left({\cos \theta + i \sin \theta}\right)}\right) }} {{eqn | l = a       | r = \dfrac {\mathrm d \left({\cos \theta + i \sin \theta}\right)} {\mathrm d \theta} \dfrac {\mathrm d \left({\operatorname{alog} \left({\cos \theta + i \sin \theta}\right)}\right)} {\mathrm d \left({\cos \theta + i \sin \theta}\right)}       | c = Chain Rule for Derivatives }} {{eqn | l = a       | r = \left({-\sin \theta + i \cos \theta}\right) \frac 1 {\cos \theta + i \sin \theta}       | c = from our assumption that $\dfrac {\mathrm d \left({\operatorname{alog} z}\right)} {\mathrm d z} = \dfrac 1 z$ }} {{end-eqn}} This last equation is true regardless of the value of $\theta$. In particular, for $\theta = 0$, we must have: : $a = i$ which proves the lemma. {{qed|lemma}} We now have established there is one function which truly deserves to be called the logarithm of complex numbers, defined as: :$\log \left({z}\right) = i \arg \left({z}\right) + \log \left\vert{z}\right\vert$ Since for any $z, z_1, z_2 \in \C, x \in \R$ it satisfies: : $\log \left({z_1 z_2}\right) = \log \left({z_1}\right) + \log \left({z_2}\right)$ : $\log \left({x}\right) = \log x$ : $\dfrac {\mathrm d \left({\log \left({z}\right)}\right)} {\mathrm d z} = \dfrac 1 z$ Let its inverse function be referred to as the '''exponential''' of complex numbers, denoted as $e^z$. If we write $z$ in its polar form: : $z = \left\vert{z}\right\vert \left({\cos \theta + i \sin \theta}\right)$ we have that: : $e^{i \theta + \log \left\vert{z}\right\vert} = \left\vert{z}\right\vert \left({\cos \theta + i \sin \theta}\right)$ Consider this equation for any number $z$ such that $\left\vert{z}\right\vert = 1$. Then: : $e^{i \theta} = \cos \theta + i \sin \theta$ {{qed}}