Holomorphic extension of function Let $\Omega$ be a simply connected domain in $\mathbb{C}$ and its boundary $\partial\Omega$ be a closed curve. Also let $D$ be a simply connected domain inside $\Omega$ and its boundary $\partial D$ be a closed curve. 
If $f$ is a linear function on $\partial\Omega$, i.e. $f(z)=az+b$ and $f$ is a holomorphic function on $\partial D$. Then, I think I can say that $f$ can be holomorphically extend into $\Omega\setminus\overline{D}$. Furthermore $f(z)$ must be also linear function $az+b$ along $\partial D$ Is this correct? or not? 
 A: In complex analysis, one usually uses the term Jordan domain for a domain whose boundary is a simple closed curve. 
The function $f(z) = az+b$, initially defined on $\partial \Omega$, admits a holomorphic   extension to $\Omega\setminus \overline{D}$, also given by the formula $az+b$. I think the main question here is whether this is the only extension. In other words, if $g$ is holomorphic in $\Omega\setminus \overline{D}$, continuous on $\overline{\Omega}\setminus \overline{D}$, and satisfies $g(z)=az+b$ for $z\in\partial \Omega$, does it follow that $f(z)=az+b$ in $\Omega\setminus \overline{D}$? 
The answer is yes. Proof. Let $h(z)=g(z)-az-b$. Our goal is to show that $h$ is identically zero. Let $F$ be  a conformal map $F:B\to\Omega$, where $B$ is the unit disk. The composition $\tilde h= h\circ F$ is holomorphic in the doubly-connected domain $B\setminus F^{-1}(\overline{D})$, and $\tilde h(z)\to 0$ as $|z|\to 1$. The domain $B\setminus F^{-1}(\overline{D})$ contains an annulus $\{z:r<|z|<1\}$ for some $r<1$. The function $\tilde h $ is represented by its  Laurent series $\sum c_n z^n$ in this annulus. For every $n$ and for every $\rho\in (r,1)$ we have 
$$c_n=\frac{1}{2\pi i}\int_{|z|=\rho} z^{-n-1}\tilde h(z)\,dz \tag1$$
Letting $\rho\to 1$ in (1) yields $c_n=0$. Thus, $\tilde h$ is identically zero in the annulus $\{z:r<|z|<1\}$. By this identity theorem for holomorphic functions, it is identically zero in $B\setminus F^{-1}(\overline{D})$.
