A holomorphic function determined by its values on two lines. There is a complex plane $\mathbb{C}$ and two different lines on it : $l,k$
We know all the values of some holomorphic function defined on that two lines.

Is this true that there is at most one function defined on  whole plane $\mathbb{C}$ so that it has the same values on that lines?

I know that in case of region on plane with nonzero surface it is determined-because the function is smooth.
I know also that one line is not enough (consider a plane croosing through a constant line but with different angle)
I tried to solve this in following way. Lets $l,k$ intersect at point $P$. Then we may consider two versors on that lines with begining at $P$. Then all the values on plane are linear combinations of that two vectors. 
I thought that it determines behaviour of the function but i need your help.
 A: Yes, it is unique. Given an open connected set $\Omega \subset \Bbb{C}$, if a function $f$ is holomorphic on $\Omega$ and vanishes on a subset of $\Omega$ that contains some of its limit points, then it's the zero function. 
So suppose you have two holomorphic functions on some open connected set containing the lines that have the same values on the two lines. Since the lines contain at least one of their limit points (any point on the line is a limit point!) then the function $h = f-g$ is holomorphic and zero on the lines. Thus, $h$ is zero everywhere else, which means $f=g$.
Note that with the aforementioned theorem, a line is indeed enough.
EDIT: About your suggestion specifically, the vector space structure of points on lines is not specifically the reason for the uniqueness of such function. The reason is differential in nature and it deals with the fact that an holomorphic function's value at a point $z_0$ is uniquely determined by an integral involving $f(z)/(z-z_0)$ around the point.
Regarding your plane example, I don't really know how you might want to define the function as a plane in the $\Bbb{R}^3 $ sense. If you're thinking of:
$$ f(x, y) = ax + by \\
\implies f(x + iy) = ax + iby$$
And thus for $z\in \Bbb{C}$:
$$f(z) = a \text{Re}(z) + ib \text{Im}(z)$$
Note that this function is, in fact, not holomorphic.
A: Prescribe the values on the lines. A holomorphic function taking those prescribed values may exist or not on a neighborhood of the lines. However, it it exists it must be unique -- if the domain is connected -- by the Identity Principle that says 

Let $U\subset \mathbb{C}$ be a connected open set and let $f,g\colon U \rightarrow \mathbb{C}$ be two holomorphic functions. Let $S= \{z\in U \mid f(z) =g(z)\} $. If $S$ has a accumulation point in $U$ then $f(z)=g(z)$ for every $z\in U$.

See any standard book on complex analysis for a proof of this result.
