Let $\Omega=\{e^{x+iy}: x\in(1,10),~ |y-x|<1\}$. Show that there exists a branch $L$ of logarithm defined on $\Omega$. Also, how many values does the difference $(\log − L)$ take, where $\log$ is the principal branch of logarithm?
I showed that $0\notin\Omega$ and $0<\arg z<\arctan 2,$ and defined $L(z)=\log r +i\theta~$ for $0<\theta <\arctan2$. This shows that $L$ is the branch on $\Omega$. Now for the difference $(\log −L)$: since $\log$ assumes arguments in $(-\pi, \pi)$, we conclude that $(\log −L)$ assumes arguments in $(-\pi-\arctan2,\pi)$. Since $2\pi<\pi-(-\pi-\arctan2)<4\pi$, the difference takes two values. Is that correct?
For a branch of the logarithm to be defined, it is enough for $\Omega$ to be simply connected and exclude $0$. As you mention, $0\notin \Omega$. To see that $\Omega$ is simply connected, note that it is the image of the restricted exponential \begin{align*} f\colon U &\longrightarrow \Omega \\ z &\longmapsto e^z, \end{align*} where $U = \{x + i y : x \in (1, 10), \lvert x - y\rvert < 1\}$ is simply connected since it is convex.
To see that $f$ is injective, suppose that $f(x + iy) = f(x' + iy')$. Then $x = x'$ and $y = y' + 2\pi k$ for some $k \in \mathbf{Z}$. But notice that, since $x + iy, x' + iy' \in U$, $$ |y - y'| = |y - x - (y' - x')| \leq |y - x| + |y' - x'| < 2, $$ which means that $k = 0$ so that $y = y'$, and so $f$ is injective. It's also surjective by the definition of $\Omega$.
Moreover, the exponential map is continuous an open, so $f$ is a homeomorphism. Thus $\Omega$ is homeomorphic to $U$ and so is simply connected. Hence, it admits a branch of the logarithm.
Given that the $e^{x + iy}$ representation of elements of $\Omega$ is unique over $x + i y \in U$, any branch of the complex logarithm over $\Omega$ must be of the form $L(e^{x + iy}) = x + i(y + 2\pi k)$ for some $k \in \mathbf{Z}$.
On the other hand, the principal value of the logarithm satisfies, for $x + iy \in U$, \begin{align*} \log(e^{x + i y}) = \begin{cases} x + i y &\text{if } y\in (0, \pi], \\ x + i(y - 2\pi) &\text{if } y\in (\pi, 3\pi],\\ x + i(y - 4\pi) &\text{if } y\in (3\pi, 11). \end{cases} \end{align*}
Therefore, we have that \begin{align*} \log(e^{x + i y}) - L(e^{x + i y}) = \begin{cases} -2\pi k i &\text{if } y\in (0, \pi], \\ -2\pi(k+1) i &\text{if } y\in (\pi, 3\pi],\\ -2\pi(k+2) i &\text{if } y\in (3\pi, 11). \end{cases} \end{align*}
That is, $\log(z) - L(z)$ takes $3$ distinct values on $\Omega$.