Why is this composite function entire? 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.)
 A: 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*}

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.
