User:Misael.G.Mx/Sandbox1

It is a consequence of the multiplication definition of complex numbers that the $\map \arg z$ function, for $z \in \C$, satisfies the relationship

which means that $\map \arg z$ 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 $\map \arg z$ can not be considered a generalization to complex values of the ordinary $\log$ function for real values, since for $z \in \R$, we have $0 = \map \arg z \ne \log z$.

If we do wish to generalize the $\log$ function to complex values, we can use $\map \arg z$ to define a set of functions
 * $\map {\operatorname {alog} } z = a \map \arg z + \log \size z$, 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:


 * $\map {\operatorname {alog} } z = a \map \arg z + \log \size z$

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.
 * $\map {\operatorname {alog} } {z_1 z_2} = \map {\operatorname {alog} } {z_1} + \map {\operatorname {alog} } {z_2}$, and
 * $\map {\operatorname {alog} } x = \log x$

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 $\map \arg x = 0$, 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 $\log$
 * $\dfrac {\d \log x} {\d x} = \dfrac 1 x$

Lemma 2
Let $\map {\operatorname {alog} } z = a \map \arg z + \log \size z$, then if:
 * $\dfrac {\map \d {\operatorname {alog} z} } {\d z} = \dfrac 1 z$

we must have
 * $a = i$.

Proof Lemma 2
Let $z \in \C$ be such that $\size z = 1$, and $\map \arg z = \theta$.

Then:
 * $z = \cos \theta + i \sin \theta$

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

We now have:
 * $a \theta = \map {\operatorname {alog} } {\cos \theta + i \sin \theta}$

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:
 * $\map {\mathbf {log} } z = i \map \arg z + \log \size z$

Since for any $z, z_1, z_2 \in \C, x \in \R$ it satisfies:
 * $\map {\mathbf {log} } {z_1 z_2} = \map {\mathbf {log} } {z_1} + \map {\mathbf {log} } {z_2}$
 * $\map {\mathbf {log} } x = \log x$
 * $\dfrac {\map \d {\map {\mathbf {log} } z} } {\d z} = \dfrac 1 z$

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

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