# Analytic continuation - How is it that $f_1(z)=f_2(z)$ in some part of the $\Omega_1 \cap \Omega_2$ but not for the rest?

I am struggling with Complex Analysis and really need some help to understand my lecture notes:

Definition: Let $$f(z)$$ be holomorphic in a domain $$\Omega$$, and let $$g(z)$$ be defined on a subset S of $$\Omega$$ that has a point of accumulation in $$\Omega$$. If $$f(z) = g(z)$$ for all $$z \in S$$, we call $$f(z)$$ the analytic continuation of g(z) to the domain $$\Omega$$.

• If $$f_1(z)$$ and $$f_2(z)$$ are both analytic continuations of $$g(z)$$ to the same domain $$\Omega$$, then $$f_1(z) = f_2(z),\ \forall z \in \Omega$$
• If $$f_1(z)$$ is an analytic continuation of $$g(z)$$ to the domain $$\Omega_1$$ and $$f_2(z)$$ is an analytic continuation of $$g(z)$$ to a different domain $$\Omega_2$$ we may have

$$\qquad f_1(z) \neq f_2(z)$$ at some points in $$\Omega_1 \cap \Omega_2$$.

We are only certain that $$f_1(z) = f_2(z), \forall z \in T$$ where $$T$$ is a domain (we emphasise connectedness of $$T$$) such that
$$\qquad S \subset T \subset \Omega_1 \cap \Omega_2$$.

I get the first bullet point but not quite for the second. How is it that $$f_1(z)=f_2(z)$$ in some part of the $$\Omega_1 \cap \Omega_2$$ but not for the rest? Can someone shed some light on this? Really appreciate the help!

How about $\Omega_1$ being the complex plane with the origin and negative real axis removed, and $\Omega_2$ being the complex plane with the origin and positive real axis removed? Let $f_1$ and $f_2$ be branches of the complex square root. Precisely, $f_1$ on $\Omega_1$ is defined by $f_1(re^{it})=\sqrt r e^{it/2}$ for $r>0$ and $-\pi<t<\pi$ and $f_2$ on $\Omega_1$ is defined by $f_2(re^{it})=\sqrt r e^{it/2}$ for $r>0$ and $0<t<2\pi$. Then $f_1=f_2$ on the upper half-plane, but $f_1=-f_2$ on the lower half-plane.
Consider $$\Omega_1=\mathbb C\setminus\{ir: r\ge 0 \}, \qquad \Omega_2=\mathbb C\setminus\{-ir: r\ge 0 \}.$$ Let $f_1,f_2$ be the logarithms on $\Omega_1,\Omega$, respectively, so that $$f_1(x)=f_2(x)\in\mathbb R, \quad x\in\mathbb R^+.$$ Then $$f_1(z)=f_2(z), \quad\text{if Re\,z>0}.$$ But, $$f_1(x)-f_2(x)=-2\pi i, \quad\text{if Re\,z<0}.$$