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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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    $\begingroup$ What exactly are you asking? The formulas for real part and imaginary part are pretty standard, your formulas for sine and cosine follow directly from Euler's formula. The formula using the logarithm is one particular definition of the logarithm. The formulas for the functions you gave below are not quite correct, since you omit the $z$ dependence of $\theta$ and $f(z)^* \ne f(z^*)$ in general. $\endgroup$ – filmor Dec 5 '13 at 9:07
  • $\begingroup$ The formula for $\sin \theta$ is also incorrect. (There's an $i$ missing in the denominator.) $\endgroup$ – mrf Dec 5 '13 at 10:21
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    $\begingroup$ @Eddy, what do you mean by "complex transformation"? What do you mean by "preserve angles"? The first set of equations is very well known and used all the time while the second set relies on a property ($f(z)^* = f(z^*)$) that you neither define nor show for the function you apply it to. $\endgroup$ – filmor Dec 5 '13 at 11:46
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    $\begingroup$ A simple concrete example to illustrate @filmor's point: Let $f(z) = z+i$. Then $f(i) = 2i$, but $f(-i) = 0$, so neither $\arg f(z^*) = -\arg f(z)$ nor $|f(z)|=|f(z^*)|$ holds in general for holomorphic functions. $\endgroup$ – mrf Dec 5 '13 at 12:05
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    $\begingroup$ @Eddy, if $f(z) = z+i$ then $f(z^*) = z^* + i$ though. $\endgroup$ – Antonio Vargas Dec 5 '13 at 18:50
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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.

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    $\begingroup$ Eddy, I read through some of your posts, and commend you for discovering so much about complex analysis with little formal training. If you want to delve deeper, I could suggest you look into any introductory textbook about complex numbers or complex analysis. I like this one for free: people.math.gatech.edu/~cain/winter99/complex.html $\endgroup$ – rajb245 Dec 29 '13 at 19:43
  • $\begingroup$ thanks you for the link! Indeed, i'm a graphist, not a mathematician... $\endgroup$ – Eddy Khemiri Dec 30 '13 at 11:28
  • $\begingroup$ but I use the formulas, and as so far, it works well. But I think that the problem is that one should prove that if $\arg(f(z))=\theta$ then $\arg(f(z^*))=-\theta$, and as many comments point out, this may only be true for real-valued fonctions (or series). but choosing $f(z)^*$ instead of $f(z^*)$ should work fine? $\endgroup$ – Eddy Khemiri Dec 30 '13 at 11:40
  • $\begingroup$ Yes $f(z)^*$ in place of $f(z^*)$ in your formulas is the most general form. This follows directly from formally substituting $z\rightarrow f(z)$. For real valued functions, you have $f(z^*)=f(z)^*=f(z)=\Re(f(z))$, which gives the formulas you provided. $\endgroup$ – rajb245 Dec 31 '13 at 18:01
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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 $f(z)=\frac12(z+z^*).$

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