# Why $r$ must near real number $1$

Here is a theorem in Chapter 3 of M.Stein's complex Analysis

Theorem 6.1 Suppose that $\Omega$ is simply connected with $1 \in \Omega$, and $0\not\in \Omega$. Then in $\Omega$ there is a branch of the logarithm $F(z)=\log_{\Omega}(z)$ so that
(i) $F$ is holomorphic in $\Omega$
(ii)$e^{F(z)}=z$ for all $z \in \Omega$
(iii)$F(r)=\log r$ whenever $r$ is a real number and near $1$.

then the author define $$\log_{\Omega}(z)=\int_{\gamma}\frac{1}{\omega}\text{d}\omega$$ where $\gamma$ is any curve in $\Omega$ connecting $1$ to $z$. then author verified (i) and (ii).

but the third property (iii) is confused me. I don't understand why $r$ must near $1$. and in the book ,the author proves as the below:

Finally, if $r$ is real and close to $1$ we can choose as a path from $1$ to $r$ a line segment on the real axis so that $$F(r)=\int_1^r\frac{\text{d}x}{x}=\log r$$ by the usual formula for the standard logarithm.

In his proof I can't seen any necessary that $r$ must near $1$. so why author say it must near $1$? thanks very much.

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Because larger $r$'s may not be in $\Omega$. In which case $F(r)$ is not defined. –  1015 Feb 21 '13 at 17:26
thank you very much. I am a fool. it must be the reason. –  Laura Feb 21 '13 at 17:28
Note that $\Omega$ is simply connected, but $\Omega\cap \mathbb R$ may be quite disconnected. The conclusion is only valid for the connected component of $\Omega\cap \mathbb R$ that contains $1$ as for all other components, any path from $1$ to $r$ may need to wind around $0$, thus possibly producing $\ln r +2ki\pi$ there with $k\ne 0$.
An example of such an $\Omega$ is as follows (sketch it!):
$$\Omega=\left\{(1+a)e^{ib}\ \bigg|\ -1< b-\frac12<a<b+\frac12\right\}.$$