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Let G=\mathbb{C}\backslash \{x \in \mathbb{R} | |x|\ge 1 \}. We want to find a holomorphic function $f$ so that $f(0)=i,\qquad\text{and}\qquad (f(z))^2 = z^2 - 1 \text{ for all }z\in G.$

VVVs work:

$(f(z))^{2} = z^{2}-1 = (z-1)(z+1)$

let l be a logarithmic function, then $f_{2} = \sqrt{z-1} = exp(\frac{1}{2}l(z-1))$ for z\in \mathbb{C} \backslash \{ x \in \mathbb{R} | x \ge + 1\} and $f_{1} = \sqrt{z+1}=exp(\frac{1}{2}l(z+1))$ for $z\in \mathbb{C}\backslash \{x \in \mathbb{R}| x\le -1\}$

and one chooses: $l= \frac{1}{2}log(x^{2}+y^{2})+iarctan(y/x)$

and this gives : $f_{1}f_{2} = \sqrt{z^{2}-1}$

attempt 2

Directly one sees also that $f(z) = \pm \sqrt{(z^{2}-1)}$ and with the condition $f(0)=i$ it follows that $f(z) = \sqrt{z^{2}-1}$


Is VVVs work correct ?

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    VVV, I wanted to let $y$ou know th$a$t there is a limit of 6 questions per day and 50 questions per 30 days. You're mem$b$er for 13 da$y$s now $a$nd already have 31 questions. So, you might want to slow down a little. See [here](http://meta.math.stackexchange.com/questions/2302/do-we-limit-how-often-someone-may-ask-a$b$out-their-homework/2303#2303) and [here](http://meta.math.stackexchange.com/questions/2464/question-limit-per-month) for further information.2011-11-07

2 Answers 2

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The trouble with attempt 2 is that $\sqrt{z^2-1}$ can't just be used without defining it, and defining it is really what the question is asking you to do. If $s$ is a positive real, we can define $\sqrt s$ to be the positive number $r$ such that $r^2=s$, but if $s$ is not a positive real, then we have no concept of "positive" (unless we define one).

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    I mean defining $f$ directly in terms of logs and exponentials instead of indirectly via $f_1$ and $f_2$.2011-11-08
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You need to define a holomorphic square root function on $G' = \mathbb{C} \setminus \{x \in \mathbb{R} \mid x \geq 0\}$. Letting $v = r\exp(i\theta)$ with $r>0$ and $0<\theta<2\pi$, define the square root of $v$ as $\sqrt{r}\exp(i\theta/2)$.