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Show that if $h$ is harmonic , then any mth order partial derivative of $h$ is a linear combination of $\dfrac{\partial^{m}h}{\partial z^{m}}$ and $\dfrac{\partial^{m}h}{\partial \overline z^{m}}$

My attempt:

If $h$ is analytic, we have $\frac{\partial^{k+l}}{\partial z^{k} \partial \overline z^{l}}=0$ unless $l=0$. the only derivative that doesnt vanish is $\frac{\partial^{m}h}{\partial z^{m}}= h^{(m)}(z)$. If $h$ is a analytic conjugate: $\frac{\partial^{k+l}h }{\partial \overline z^{k} \partial z^{l}}=0$ unless $l=0$. If $h$ is harmonic then $h=$analytic+ conjugate analytic, then all the derivatives vanish except for $\frac{\partial^{k}h}{\partial z^{k}}=\frac{\partial^{l}h}{\partial \overline z^{l}}$

Can you tell me if I am right? If not can you help me fix it?

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Your approach is reasonable, but the argument needs to be more precise. I would begin by noting that every first-order partial of $h$ is a linear combination of $\partial h/\partial z$ and $\partial h/\partial \bar z$. It follows that every partial of order $m$ is a linear combination of $$ \frac{\partial^m h}{\partial^k z\partial^{j}\bar z},\quad k+j=m \tag{1}$$ because the $m$-fold composition of things like $a\,\partial /\partial z + b \,\partial /\partial \bar z$ consists of derivatives like $(1)$.

The Laplace equation is expressed in complex notation as $$ \frac{\partial^2 h}{\partial z\partial\bar z}=0$$ Thus, the terms in $(1)$ where both $k$ and $j$ are positive are all zero. We are left with just $k=m, j=0$ and $k=0,j=m$.

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