# Homogeneity of translated polynomial

I am currently trying to understand the very basics of complex algebraic curves and I came across the following statement in the book by F. Kirwan (Definition 2.9):

The multiplicity of the complex algebraic curve $C$ defined by the polynomial $P(x,y)$ at a point $(a,b) \in C$ is the smalles positive integer $m$ such that $$\frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \ne 0$$ for some $i \ge 0$, $j \ge 0$ such that $i + j = m$.

The polynomial $$\sum_{i+j = m} \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \frac{(x-a)^i(y-b)^j}{i!j!}$$ is then homogeneous of degree $m$. Here is where I get stuck - it is clear that it is homgeneous in the variables $X = x - a$, $Y = y-b$, but how can I see that $$\sum_{i+j = m} \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \frac{(\lambda x-a)^i(\lambda y-b)^j}{i!j!} = \lambda^m \sum_{i+j = m} \frac{\partial^m P}{\partial x^i \partial y^j}(a,b) \frac{(x-a)^i(y-b)^j}{i!j!} \quad ?$$

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