If $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ holds for $0In Special Functions p. 10, it has proven that
$$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)},$$
for $0<z<1$. Then it says that this equality implies for $0<\operatorname{Re}(z)<1$.
I do not understand this implication. How do we know that it's true in the imaginary part?
 A: This is due to the identity theorem for holomorphic functions:

Let $U \subset \mathbb{C}$ open and connected, and $f, g \colon U \to \mathbb{C}$ holomorphic. If the set $E := \{ z \in U : f(z) = g(z)\}$ has an accumulation point in $U$, then $E = U$.

Remark: It is essential that the set $E$ has an accumulation point in $U$. If $E$ only has accumulation points at the boundary of $U$, then the conclusion need not hold.
Since every point of the interval $I = (0,1)$ is an accumulation point of $I$, the identity
$$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin (\pi z)}\tag{1}$$
for $z \in I$ extends to every connected open set $U$ with $I \subset U$ (actually, $I \cap U \neq \varnothing$ suffices) on which both sides of $(1)$ are holomorphic functions by the identity theorem.
With the integral representation
$$\Gamma(z) = \int_0^{+\infty} t^{z-1} e^{-t}\,dt,\tag{2}$$
which is valid for $\operatorname{Re} z > 0$, one gets the holomorphic function $z \mapsto \Gamma(z)\Gamma(1 - z)$ on the strip $A = \{ z \in \mathbb{C} : 0 < \operatorname{Re} z < 1\}$. Clearly $z \mapsto \dfrac{\pi}{\sin (\pi z)}$ is holomorphic on $\mathbb{C}\setminus \mathbb{Z} \supset A$, and $A$ is connected, so this yields $(1)$ for all $z \in A$.
If one has a representation of the $\Gamma$-function valid for $z \in \mathbb{C}\setminus \{ -n : n \in \mathbb{N}\}$, one sees that $z \mapsto \Gamma(z) \Gamma(1 - z)$ is in fact holomorphic on $\mathbb{C}\setminus \mathbb{Z}$, and hence the identity $(1)$ holds on the connected open set $\mathbb{C}\setminus \mathbb{Z}$.
