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Let $f:D\subsetℂ\to \mathbb{R}$ be a non-analytic function. I know that $f$ is an odd function on $D$. The function $f$ is the argument (angle) of an analytic function. $f$ has no known form. Be odd function is all I know about it

Is it true that $D=\mathbb{C}$, i.e., an odd function in $D$ can be extended to all of the complex plane?

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Is there any reason to expect that it must be? –  Adam Saltz Jan 1 '13 at 14:56
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Since I have a function that verify this property. –  ZE1 Jan 1 '13 at 14:58
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What property? You have an odd function defined on a set and that function can be extended to the whole complex plane? [As written, your question is very confusing. Are $f$ and $h$ supposed to be the same function? Is $D$ supposed to be a domain or the unit disk? By "must be defined" do you mean "can be extended?"] –  Adam Saltz Jan 1 '13 at 15:05
    
@ Adam Saltz: Yes, you are right. I have edited the question. –  ZE1 Jan 1 '13 at 15:10
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The function $f$ is the argument (angle) of an analytic function. $f$ has no known form. Be odd function is all I know about it. –  ZE1 Jan 1 '13 at 15:25

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No. For example, let $$F(z)=\exp\left(\frac{1}{1-z}-\frac{1}{1+z}\right)$$ The argument of $F$ is odd on the unit disk $D$ because $F(-z)=1/F(z)$ for $z\in D$. Yet, there is no natural extension of $\arg F$ to $\mathbb C$ since $F$ has essential singularities at $\pm 1$.

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