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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 you know that there is a limit of 6 questions per day and 50 questions per 30 days. You're member for 13 days now and already have 31 questions. So, you might want to slow down a little. See here and here for further information. –  t.b. Nov 7 '11 at 20:03

2 Answers 2

up vote 1 down vote accepted

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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Edited version correct?? –  VVV Nov 7 '11 at 22:37
    
Could be, but why introduce two functions $f_1$ and $f_2$ instead of doing it all in one go? –  Gerry Myerson Nov 8 '11 at 0:09
    
What do you mean by doing it all in one go?? –  VVV Nov 8 '11 at 8:49
    
I mean defining $f$ directly in terms of logs and exponentials instead of indirectly via $f_1$ and $f_2$. –  Gerry Myerson Nov 8 '11 at 12:10

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 $\sqrt{r}\exp(i\theta/2)$.

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