Consider the function on the variable $x\in R$ and the parameter $a\in R$:
$$f(x)=\frac{a-\sqrt{x}}{a+\sqrt{x}}$$
Of course if $x>=0$ you have a simple function $f:R \rightarrow R$.
If $x<0$ you can write:
$$f(x)=\frac{a-\sqrt{(-1)(-x)}}{a+\sqrt{(-1)(-x)}}$$
$$f(x)=\frac{a-[\pm i\sqrt{(-x)}]}{a+[\pm i \sqrt{(-x)}]}$$
So now there are 4 possibilities according to the signs you put:
$$\binom{+}{+} \rightarrow f(x)=\frac{a-[i\sqrt{(-x)}]}{a+[i \sqrt{(-x)}]}=e^{i\theta(x)} $$
$$\binom{+}{-} \rightarrow f(x)=\frac{a-[+i\sqrt{(-x)}]}{a+[-i \sqrt{(-x)}]}=1 $$
$$\binom{-}{+} \rightarrow f(x)=\frac{a-[-i\sqrt{(-x)}]}{a+[+i \sqrt{(-x)}]} =1$$
$$\binom{-}{-} \rightarrow f(x)=\frac{a-[-i\sqrt{(-x)}]}{a+[-i \sqrt{(-x)}]} e^{-i\theta(x)} $$
Finally:
$$ f(x)=\begin{cases}
1 \\
e^{-i\theta(x)}\\
e^{i\theta(x)}
\end{cases}
$$
So I'd say that for $x<0$ $f(x)$ isn't anymore a function...Is it right?
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Landau
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This is not a function over the complex numbers, since the square root of a complex number is not a well-defined notion. See How do I get the square root of a complex number?
Hans Hüttel
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I know how to get the root of a cn, however do you confirm that my result is right? – Landau Dec 29 '18 at 20:37