Why is this composite function entire?

Question

Let $f: \mathbb{C}^n \to \mathbb{C}$ be an entire function and let $z,w \in \mathbb{C}^n$ be fixed. Then, why is $g: \mathbb{C} \to \mathbb{C}$ with $g(\lambda) = f(z + \lambda w)$ is entire? (By $f$ being entire, it means $f$ is holomorphic in each variable separately.)

Answer 1

Write $\lambda=s+it$ and $w=a_1e_1+\cdots+a_ne_n$.

Since

$$ \frac{\partial g}{\partial s}(\lambda)=\nabla_z f(z+\lambda w)\cdot(a_je_j)_j $$ $$ \frac{\partial g}{\partial t}(\lambda)=\nabla_z f(z+\lambda w)\cdot(ia_je_j)_j $$ you get \begin{align*} \frac{\partial g}{\partial \overline\lambda}(\lambda) &=\frac12 \left(\frac{\partial g}{\partial s}(\lambda) +i\frac{\partial g}{\partial t}(\lambda)\right)\\ &=\frac12 \nabla_z f(z+\lambda w)\cdot \left((a_je_j)_j+ i(ia_je_j)_j\right)=0. \end{align*}

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Notice that $$ \nabla f(\zeta) =\sum_{j=1}^n\frac{\partial f}{\partial \zeta_j}(\zeta)d\zeta_j +\sum_{j=1}^n\frac{\partial f}{\partial \overline\zeta_j}(\zeta)d\overline\zeta_j=\nabla_z f(\zeta)+\nabla_{\overline z}(\zeta), $$ thus the full writing of $\frac{\partial g}{\partial s}(\lambda)$ for example is (here $\zeta$ is the variable in $\Bbb C^n$) $$ \sum_{j=1}^n\frac{\partial f}{\partial \zeta_j}(z+\lambda w)a_je_j +\sum_{j=1}^n\frac{\partial f}{\partial \overline\zeta_j}(z+\lambda w)\overline a_je_j $$ which makes no difference since $\frac{\partial f}{\partial \overline\zeta_j}=0$ being $f$ holomorphic.