simple tools to extract Re,Im,Abs… of any complex function

I’ve developped my own set of simple yet powerful tools to work on complex functions. I would like to know if these simple tools are currently used in complex analysis.

Let’s $z = x + i y = |z| e^{i\theta}$ and it’s complex conjuguate $z^* = x – i y = |z| e^{-i\theta}$, where $\theta = \text{Arctan}(y/x)$, we can write:

$
\frac{1}{2i}\text{Log}\left( \frac{z}{z^*} \right) =\text{ }\frac{1}{2i}\text{Log}\left( \frac{|z|e^{i \theta }}{|z|e^{-i \theta }} \right) = \theta =\arg(z)
$

$
\sqrt{z z^*}=\sqrt{ |z|e^{i \theta }|z|e^{-i \theta }}=|z|=\text{Abs}(z)
$

$
\frac{1}{2}( z + z^*)=|z|\frac{(e^{i \theta }+e^{-i \theta })}{2}= \mathfrak{Re}(z)
$

$
\frac{1}{2}( \frac{z + z^*}{|z|})=\frac{(e^{i \theta }+e^{-i \theta })}{2}=\text{Cos}(\theta )
$

$
\frac{1}{2i}( z – z^*)=|z|\frac{(e^{i \theta }-e^{-i \theta })}{2i}=\mathfrak{Im}(z)
$

$
\frac{1}{2i}( \frac{z – z^*}{|z|})=\frac{(e^{i \theta }-e^{-i \theta })}{2i}=\text{Sin}(\theta )
$

Let’s now $f(z)$ be any complex function or composition of functions $g(…f(z))$, the same results hold when plugging $f(z)$ and $f(z^*)$ instead of $z$ and $z^*$, giving the tools set:

$
\frac{1}{2i}\text{Log}\left( \frac{f(z) }{f(z^*)} \right) =\text{ }\frac{1}{2i}\text{Log}\left( \frac{|f(z)|e^{i \theta }}{|f(z)|e^{-i \theta }} \right) = \theta =\arg(f(z))
$

$
\sqrt{f(z) f(z^*)}=\sqrt{ |f(z)|e^{i \theta }|f(z)|e^{-i \theta }}=|f(z)|=\text{Abs}(f(z))
$

$
\frac{1}{2}\left( f(z)+f(z^*) \right)=|f(z)|\frac{\left(e^{i \theta }+e^{-i \theta }\right)}{2}= \mathfrak{Re}(f(z))
$

$
\frac{1}{2}\left( \frac{f(z)+f(z^*)}{|f(z)|}\right)=\frac{\left(e^{i \theta }+e^{-i \theta }\right)}{2}=\text{Cos}(\theta )
$

$
\frac{1}{2i}\left( f(z)-f(z^*)\right)=|f(z)|\frac{\left(e^{i \theta }-e^{-i \theta }\right)}{2i}=\mathfrak{Im}(f(z))
$

$
\frac{1}{2i}\left( \frac{f(z)-f( z^*)}{|f(z)|}\right)=\frac{\left(e^{i \theta }-e^{-i \theta }\right)}{2i}=\text{Sin}(\theta )
$

So, are these simple formulas using $f(z)$ and $f(z^*)$ to extract the complex components of functions known? Is there in complex analysis any litterature about?

EDIT: after comments, may be we should consider $f(z)^*$ instead of $f(z^*)$ …?

Here are some examples showing their use.

Using the formula for the Zeta function $\zeta (\mathit{z})=\prod_{k=1}^{\infty } \frac{1}{1-p_k{}^{-\mathit{z}}}$ for $z>1$, where $p_k$ is the $k^{th}$ prime number, we can quite easily find:
$$
\sqrt{\zeta (z) \zeta(z^*)}=|\zeta (\mathit{z})|=\prod_{k=1}^{\infty } \sqrt{\frac{p_k{}^{2x}}{p_k{}^{2x}-2p_k{}^x\cos \left(y \log \left(p_k\right)\right)+1}}
$$

$$
\frac{1}{2i}\text{Log}\left( \frac{\zeta(z) }{\zeta( z^*)} \right) = \arg( \zeta (z) )=-\sum_ {k = 1}^{\infty}\sum _ {q = 1}^{\infty}\frac {1} {k p_q^{k x}}\text {Sin}( k y \text{ Log}(p_q ) )
$$

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So, are these simple formulas using $f(z)$ and $f(z^*)$ to extract the
complex components of functions known? Is there in complex analysis
any litterature (sic) about?

Yes. When you have provided correct formulas, they are standard identities that can be found in introductory texts about complex numbers. You have written them in a form applicable to functions instead of complex numbers, and I believe you are asking if this is a heretofore unknown way of writing formulas for complex functions. However, I think you should note that note that for a fixed $z\in \mathbb{C}$, $f(z)$ is just another number in $\mathbb{C}$. If a formula holds for an arbitrary complex number $z$, then since $f(z)$ is also a complex number, the formula will hold for that number too. You have applied known identities to about complex numbers to complex functions, but $f$ maps $\mathbb C$ to itself, so you’re still talking about known identities on $\mathbb C$.

To see this, think of defining a new complex variable, $y=f(z)$. Now rewrite all the functional identities you presented in this new variable (i.e. replace $f(z)$ with $y$ wherever you see it), and you’ve recovered the original, known identities, except you’ve replaced $z$ with $y$. What’s the fundamental difference between $y$ and $z$? They both represent arbitrary numbers in the complex plane. The point is that since $f(z)\in\mathbb C$, $y\in\mathbb C$, and $z\in\mathbb C$, you can write the identities, which are properties of any numbers in $\mathbb C$, in terms of any of these you like. All these representations are valid. Sets of formulas in terms of complex functions may not appear explicitly in the literature, but because complex functions are complex numbers for all their arguments, you can freely replace $z$ by $f(z)$ in the formulas. They both represent arbitrary numbers in $\mathbb C$ and the formulas naturally still hold.

You will, in fact, need to use $f(z)^*,$ instead, as pointed out in the comments, since (in general) $f(z^*)$ need not be identically equal to $f(z)^*$.

In point of fact, if $f:\Bbb C\to\Bbb C,$ then one necessary condition to have $f(z^*)=f(z)^*$ is that $f$ be a real-valued function when restricted to the reals. Indeed, if there is some real $r$ such that $f(r)=w$ for some $w\in\Bbb C\setminus\Bbb R,$ then $$f(r^*)=f(r)=w\ne w^*=f(r)^*.$$

Antonio Vargas points out that an entire function with all Maclaurin coefficients real satisfies the desired property, but there are other functions that do the trick and aren’t even holomorphic, such as $f(z)=\frac12(z+z^*).$