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Note:

In this question, a complex number is counted as a vector initiated from the origin.

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Is there a holomorphic function $B:\mathbb{C}^2 \to \mathbb{C}$ such that for every two non zero complex numbers $z,w$ with $z/w \notin \mathbb{R},$ the vector $B(z,w)$ is a non zero vector indicating to the direction of the bisector of the angle $\angle (z,w) $?

Motivation:

The initial formula for the "Bisector" of $\angle (z,w) $ is $B'(z,w)=|z|w+|w|z$. But it is not a holomorphic function.(It is not even smooth at $z=0$ or $w=0$). So we search for a holomorphic remedy, a holomorphic function $B$ defined on whole $\mathbb{C}^2$ such that $B(z,w)$ is real proportional to $(|z|w+|w|z)$ via a non constant real function $\lambda$.

What about if we require that such $\lambda $ be positive(Non negative)?

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2 Answers 2

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Here is a better answer than the other answer I gave, which is currently accepted. It also answers some of your questions in the comments on that answer. Pick a branch of log, then $B$ and $ (zw)^{1/2}$ are holomorphic and their arguments differ by a multiple of $\pi$ wherever they are both defined. So their quotient is a real holomorphic function and hence a real constant wherever it is defined. So locally, $B$ has to be a real multiple of some branch of $ (zw)^{1/2}$.

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    $\begingroup$ Note that answers aren't threaded, and that even their physical position may change (so "better than the previous answer", which some people try, doesn't work either). To avoid appearing (as it did to me originally) like you are favourably comparing your own answer to someone else's, you may want to point out that you are saying this is "a better answer" than your own answer (currently accepted), at mathoverflow.net/a/292636/2383 . $\endgroup$
    – LSpice
    Commented Feb 10, 2018 at 22:36
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    $\begingroup$ Ah, I had not thought about how that phrase read, thanks. $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 11, 2018 at 8:09
  • $\begingroup$ "Previous" - would people on MO read that as about spatial position of answers? I would read it as about time of posting. But maybe that is too ambiguous because of editing? $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 11, 2018 at 8:16
  • $\begingroup$ @LSpice I changed the accepted answer. Thank you for your comment. $\endgroup$
    – Ali Taghavi
    Commented Feb 11, 2018 at 11:04
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No. Let $w=1$ for simplicity. If the imaginary part $\Im z> 0$ then $\Im B(z,1)>0$ and if $\Im z< 0$ then $\Im B(z,1)<0$, so $B(z,1)$ vanishes on the line $\Im z=0$ so it is uniformly 0. The same thing happens for any $w$, just replace $\Im z$ by $\Im (z/w)$.

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  • $\begingroup$ Ah, this only works if $\lambda $ is positive, i did not see that part of the question. $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 10, 2018 at 10:51
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    $\begingroup$ OK, restrict $z$ to a small disc where $\lambda$ has a fixed sign, without loss of generality it is positive, then apply the argument I gave. $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 10, 2018 at 10:55
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    $\begingroup$ Sure, just pick a branch of the logarithm and let $B=(zw)^{1/2}$. $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 10, 2018 at 11:02
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    $\begingroup$ @AliTaghavi, take $B= w (z/w)^{1/2}$. This is in effect another way of saying $B=(zw)^{1/2}$, but perhaps writing it this way makes it clearer that it can be defined on your domain! $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 12, 2018 at 13:04
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    $\begingroup$ @AliTaghavi I think you should post that as a new question, making it clearer what you mean. It seems like you have a non-holomorphic function $F(A,C,D)$ of three complex variables $A$, $C$, $D$, and you want to find a set $S\subset \mathbb{C}^3$ on which it coincides with a holomorphic function $H(A,C,D)$. I speculate that you should think carefully about what properties you want $S$ to have. Does it have to be a complex manifold, for example? $\endgroup$
    – Simon L Rydin Myerson
    Commented Feb 13, 2018 at 14:37

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