User:Misael.G.Mx/Sandbox1

It is a consequence of the multiplication definition of complex numbers that the $\arg \left({z}\right)$ function, for $z \in \C$, satisfies the relationship

Which means that $\arg \left({z}\right)$ is a kind of logarithm, in the sense that it satisfies the fundamental property of logarithms: $\log{xy}=\log{x}+\log{y}$.

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 $z \in \R$, we have $0=\arg{(z)} \ne \log{z}$.

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
 * $\operatorname{alog} \left({z}\right) = a\arg\left({z}\right) + \log\left|{z}\right|$, for any $a \in \C$

which 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 any $a,z\in\C$, define the (complex valued) function $\operatorname{alog}$ as
 * $\operatorname{alog} \left({z}\right) = a\arg\left({z}\right) + \log\left|{z}\right|$

then, for any $z_1,z_2\in\C$, and $x\in\R$ we have This means that our (complex valued) $\operatorname{alog}$ functions can genuinely be considered generalizations of the (real valued) $\log$ function.
 * $\operatorname{alog} \left({z_1z_2}\right) = \operatorname{alog} \left({z_1}\right) + \operatorname{alog} \left({z_2}\right)$, and
 * $\operatorname{alog} \left({x}\right) =\log{x}$

Proof of Lemma 1
Let $z_1,z_2$ be any 2 complex numbers, straightforward substitution on the definition of $\operatorname{alog}$ yields:

Second part of our lemma is even more straightforward since for $x\in\R$, we have $\arg\left(x\right)=0$, then

Which concludes the proof of Lemma 1.

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:
 * $\dfrac{\mathrm{d}({\operatorname{alog}{z}})}{\mathrm{d}{z}}=\dfrac{1}{z}$,

we must have
 * $a=i$.

Proof Lemma 2
Let $z\in\C$ be such that $\left\vert z \right\vert = 1$, and $\arg{(z)}=\theta$ then
 * $z=\cos{\theta}+i\sin{\theta}$

Plugging those values in our definition of $\operatorname{alog}$:

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

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.

Now we've established there is one function which truly deserves to be called the logarithm of complex numbers, defined as:
 * ${\bf log}(z) = i\arg\left({z}\right) + \log\left|{z}\right| $

Since for any $z,z_1,z_2\in\C,x\in\R$ it satisfies:
 * ${\bf log}(z_1z_2) = {\bf log}(z_1)+{\bf log}(z_2)$
 * ${\bf log}(x) = \log{x}$
 * $\dfrac{\mathrm{d}[{\bf log}(z)]} {\mathrm{d}{z}}=\dfrac{1}{z}$

Lets call it's inverse function the exponential of complex numbers, denoted as $e^z$, then if $z=|z|(\cos{\theta}+i\sin{\theta})$
 * $e^{i\theta + \log\left|{z}\right|}=|z|(\cos{\theta}+i\sin{\theta})$

which implies:
 * $e^{i\theta}=\cos{\theta}+i\sin{\theta}$