Prove that $\bar{P}_{\bar{z}}=P_z,\ (P_z,P_z)=(P_{\bar{z}},P_{\bar{z}})$ with $P_z=\dfrac{\partial{P}}{\partial{z}}$

Question

I have a problem:

For $P$ is a nonzero real valued homogeneous polynomial of degree $k$: $$P(z,\bar{z})=\sum_{j=1}^{k-1}a_jz^j\bar{z}^{k-j}$$ where $a_j \in \Bbb C,\ a_j=\bar{a}_{k-j}$.

For $Q=\sum_{j=1}^{k-2}\alpha_jz^j\bar{z}^{k-j-1},\ Q=\sum_{j=1}^{k-2}\beta_jz^j\bar{z}^{k-j-1}$ then $$(Q,S)=\sum_{j=1}^{k-2}\alpha_j \bar{\beta}_j$$ (scalar product on the vector space of homogeneous polynomials of degree $k-1$ without a harmonic term.)

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In my book:

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I have 2 questions:

1. I think that we need $\alpha - \beta \ne 0$ (not $|\alpha| \ne |\beta|$). Bcz: $$\left\{\begin{matrix} \alpha f_0+\beta \bar{f}_0=\gamma\\ f_0+\bar{f}_0=2\text{Re}f_0 \end{matrix}\right. \implies \underset{\ne 0}{\underbrace{(\alpha-\beta)}}f_0=\gamma-2\beta \text{Re}f_0$$

2. I don't understand Why do we have $\bar{P}_{\bar{z}}=P_z$, $(P_z,P_z)=(P_{\bar{z}},P_{\bar{z}})$ where $P_z=\dfrac{\partial{P}}{\partial{z}}$.

   • $\bar{P}_{\bar{z}}=P_z$??? Why? I think that:

\begin{align*} P_z=\dfrac{\partial P}{\partial z}=\dfrac{\partial }{\partial z}\left (\sum_{j=1}^{k-1}a_jz^j\bar{z}^{k-j} \right ) &=\sum_{j=1}^{k-1}(ja_j)z^{j-1}\bar{z}^{k-j}\\ {\bar P}_{\bar z}=\dfrac{\partial \bar P}{\partial \bar z}=\dfrac{\partial }{\partial \bar z}\left (\sum_{j=1}^{k-1}\bar{a_j}\bar{z^j}{z}^{k-j} \right ) &=\sum_{j=1}^{k-1}(j\bar{a_j})\bar{z}^{j-1}{z}^{k-j} \end{align*}

Is that correct?????

Any help will be appreciated! Thanks!

Answer 1

For any fixed $\alpha,\beta\in\mathbb{C}$, the map $z \mapsto \alpha z + \beta\overline{z}$ is $\mathbb{R}$-linear. If we look at its matrix in real coordinates, we find

$$\begin{pmatrix}\operatorname{Re}\alpha + \operatorname{Re}\beta & \operatorname{Im}\beta - \operatorname{Im}\alpha\\ \operatorname{Im}\alpha + \operatorname{Im}\beta &\operatorname{Re}\alpha -\operatorname{Re}\beta \end{pmatrix},$$

whose determinant is $(\operatorname{Re}\alpha)^2 - (\operatorname{Re}\beta)^2 - \bigl((\operatorname{Im}\beta)^2 - (\operatorname{Im}\alpha)^2\bigr) = \lvert\alpha\rvert^2-\lvert\beta\rvert^2$. So the map is bijective if and only if $\lvert \alpha\rvert \neq\lvert\beta\rvert$.

If $\lvert\alpha \rvert = \lvert\beta\rvert$, but $\alpha\neq\beta$, multiplying the entire equation with a suitable $e^{i\varphi}$ brings it into the form $$\delta f_0 + \overline{\delta}\overline{f}_0 = \tilde{\gamma},$$

which only has solutions if $\tilde{\gamma}\in\mathbb{R}$, and in that case, the solution is not unique, since adding any real multiple of $i\overline{\delta}$ to $f_0$ doesn't change the left hand side.

Concerning the second question, generally, we have

$$\frac{\partial \overline{g}}{\partial\overline{z}} = \overline{\frac{\partial g}{\partial z}}$$

for any real-differentiable function $g$. In our case, $P$ is supposed to be real-valued, and hence

$$\frac{\partial\overline{P}}{\partial\overline{z}} = \frac{\partial P}{\partial\overline{z}} = \overline{\frac{\partial P}{\partial z}},$$

or, equivalently

$$\overline{\frac{\partial P}{\partial\overline{z}}} = \frac{\partial P}{\partial z},$$

which we can write as

$$\overline{P_{\overline{z}}} = P_z.$$

I think that was the intention, and the

$$\bar{P_{\bar{z}}} = P_z$$

is just a typographical mistake arising from the use of \bar{P_{\bar{z}}} instead of the proper \overline.