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Let $f:\mathbb{C^n}\rightarrow \mathbb{C}$ be a function. Define $g:\mathbb{D}\rightarrow \mathbb{C}$, by $g(t)=f(tz)$ ($\mathbb{D}$ is the unit disk in $\mathbb{C}$). Calculate $g^k(t)$ for some $k$ including terms of the form $\displaystyle \frac{\partial^n(f(tz))}{\partial^n(z)} $ for $n\leq k$.

I tried the simplest one first, i.e. when $k=1$. Here, $z=(z_1,z_2,...,z_n)$. So,by chain rule $\displaystyle g'(t)=\frac{\partial(f(tz))}{\partial(tz)}.z=\frac{\partial(f(tz))}{\partial(z)}.\frac{\partial(z)}{\partial(tz)}.z=\frac{\partial(f(tz))}{\partial(z)}.\frac{z}{t}$.

Similarly I get $\displaystyle g''(t)=\frac{\partial(f(tz))}{\partial(z)}(\frac{-z}{t^2})+\frac{\partial^2(f(tz))}{\partial^2(z)}(\frac{z^2}{t^2}).$

Is my calculations correct?

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I assume you mean $f \colon \mathbb{C}^n \to \mathbb{C}$ is holomorphic, otherwise you'd also need the $\bar{z}$ derivatives. However, the main thing however is that $z$ is a vector, you cannot just differentiate with it. You have to use the multivariable chain rule.

$\displaystyle g'(t) = \frac{\partial}{\partial t} \Big[ f(tz) \Big] = \sum_{j=1}^n \frac{\partial f}{\partial z_j}(tz) \, \left( \frac{\partial}{\partial t} \Big[ tz_j \Big] \right) = \sum_{j=1}^n \left( \frac{\partial f}{\partial z_j}(tz) \right) \, z_j = \nabla f \Big|_{tz} \cdot z $

where $\cdot$ is the dot product.

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  • $\begingroup$ I think you are missing terms like $\frac{\partial z_j}{\partial tz_j}$ $\endgroup$
    – Extremal
    Feb 9, 2017 at 15:15
  • $\begingroup$ $f$ is a function of $z$. If you wish to not abuse the letter $z$, you could assume $f$ is a function of $w = (w_1,\ldots,w_n)$. Then $\displaystyle g'(t) = \frac{\partial}{\partial t} \Big[ f(tz) \Big] = \sum_{j=1}^n \frac{\partial f}{\partial w_j}(tz) \, \left( \frac{\partial}{\partial t} \Big[ tz_j \Big] \right) = \sum_{j=1}^n \left( \frac{\partial f}{\partial w_j}(tz) \right) \, z_j = \nabla f \Big|_{w=tz} \cdot z$ $\endgroup$
    – Jiri Lebl
    Feb 10, 2017 at 5:52

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