Euler's Formula/Real Domain/Proof 3

From ProofWiki
Jump to navigation Jump to search



Theorem

Let $\theta \in \R$ be a real number.

Then:

$e^{i \theta} = \cos \theta + i \sin \theta$


Proof

It follows from Argument of Product equals Sum of Arguments that the $\map \arg z$ function for all $z \in \C$ satisfies the relationship:

$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2}$

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 $x \in \R$, we have:

$0 = \map \arg x \ne \log x$

If we do wish to generalize the $\log$ function to complex values, we can use $\arg z$ to define a set of functions:

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

for any $a \in \C$, where $\cmod z$ 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:

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

then, for any $z_1, z_2 \in \C$ and $x \in \R$:

$\map {\operatorname{alog} } {z_1 z_2} = \map {\operatorname{alog} } {z_1} + \map {\operatorname{alog} } {z_2}$

and:

$\map {\operatorname{alog} } x = \log x$

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:

\(\ds \map {\operatorname{alog} } {z_1 z_2}\) \(=\) \(\ds a \map \arg {z_1 z_2} + \log \cmod {z_1 z_2}\)
\(\ds \) \(=\) \(\ds a \paren {\map \arg {z_1} + \map \arg {z_1} } + \map \log {\cmod {z_1} \cmod {z_2} }\)
\(\ds \) \(=\) \(\ds a \paren {\map \arg {z_1} + \map \arg {z_1} } + \log \cmod {z_1} + \log \cmod {z_2}\)
\(\ds \) \(=\) \(\ds a \map \arg {z_1} + \log \cmod {z_1} + a \map \arg {z_2} + \log \cmod {z_2}\)
\(\ds \) \(=\) \(\ds \map {\operatorname{alog} } {z_1} + \map {\operatorname{alog} } {z_1}\)


Second part of our lemma is even more straightforward since for $x \in \R$, we have:

$\map \arg x = 0$

Then:

\(\ds \map {\operatorname{alog} } x\) \(=\) \(\ds a \map \arg x + \log \cmod x\)
\(\ds \) \(=\) \(\ds \log x\)

which concludes the proof of Lemma 1.

$\Box$


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 $\map {\operatorname{alog} } z = a \map \arg z + \log \cmod z$.

Then if:

$\dfrac {\map \d {\operatorname{alog} z} } {\d z} = \dfrac 1 z$

we must have:

$a = i$


Proof of Lemma 2

Let $z \in \C$ be such that:

$\cmod z = 1$

and:

$\map \arg z = \theta$


Then:

$z = \cos \theta + i \sin \theta$

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

\(\ds \map {\operatorname{alog} } z\) \(=\) \(\ds a \map \arg {\cos \theta + i \sin \theta} + \log \cmod z\)
\(\ds \) \(=\) \(\ds a \theta + \log 1 = a \theta\)

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:

\(\ds \map {\frac \d {\d \theta} } {a \theta}\) \(=\) \(\ds \map {\frac \d {\d \theta} } {\map {\operatorname{alog} } {\cos \theta + i \sin \theta} }\)
\(\ds a\) \(=\) \(\ds \dfrac {\map \d {\cos \theta + i \sin \theta} } {\d \theta} \dfrac {\map \d {\map {\operatorname{alog} } {\cos \theta + i \sin \theta} } } {\map \d {\cos \theta + i \sin \theta} }\) Chain Rule for Derivatives
\(\ds a\) \(=\) \(\ds \paren {-\sin \theta + i \cos \theta} \frac 1 {\cos \theta + i \sin \theta}\) by hypothesis: $\dfrac {\map \d {\operatorname{alog} z} } {\d z} = \dfrac 1 z$

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.

$\Box$


We now have established there is one function which truly deserves to be called the logarithm of complex numbers, defined as:

$\log z = i \map \arg z + \log \cmod z$

Since for any $z, z_1, z_2 \in \C, x \in \R$ it satisfies:

$\map \log {z_1 z_2} = \log z_1 + \log z_2$
$\log x = \log x$
$\dfrac {\map \d {\log z} } {\d z} = \dfrac 1 z$

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:

$z = \cmod z \paren {\cos \theta + i \sin \theta}$

we have that:

$e^{i \theta + \log \cmod z} = \cmod z \paren {\cos \theta + i \sin \theta}$

Consider this equation for any number $z$ such that $\cmod z = 1$.

Then:

$e^{i \theta} = \cos \theta + i \sin \theta$

$\blacksquare$