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Complex functions and inner product $\langle \frac{\partial f}{\partial z} , g\rangle $

I'm working through this academic paper : stability of the soljutions of differential equations whose author is Bernard Beauzamy. [A link to paper](https://typeset.io/pdf/stability-of-the-solutions-of-differential-equations-2byjzvdtk5.pdf)
~~In the academic paper, it work with the next norm~~
\begin{equation*}
\left\Vert f\right\Vert= \left( \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r^2} |f(re^{i\theta})|^2 r dr\frac{d\theta}{\pi} \right)^{1/2}
\end{equation*}
for analytic polynomials
and the scalar product associated is
\begin{equation*}
<f,g>= \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r^2}f(re^{i\theta})\overline{g(re^{i\theta})} r dr\frac{d\theta}{\pi}
\end{equation*}
~~The lemma that i'm trying to check say the following~~
~~For any f,g and $\mathcal{P}_2$~~
\begin{equation*}
~~\langle \frac{\partial f}{\partial z} , g angle =\langle f, zg angle \qquad \langle zf,g angle =\langle f, \frac{\partial g}{\partial z} angle~~
\end{equation*}
~~the functions f and g are given by~~
\begin{equation*}
f(z) = \sum_{j=0}^{\infty} a_j z^j, \quad g(z) = \sum_{j=0}^{\infty} b_j z^j.
\end{equation*}
~~The derivative of f with respect to z is~~
\begin{equation*}
\frac{\partial f}{\partial z} = \sum_{j=1}^{\infty} a_j j z^{j-1}.
\end{equation*}
so the scalar product is
\begin{align*}
\langle \frac{\partial f}{\partial z}, g angle &= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \left(\sum_{j=1}^{\infty} a_j j (re^{i\theta})^{j-1} ight) \overline{\left(\sum_{k=0}^{\infty} b_k (re^{i\theta})^k ight)} r \, dr \, \frac{d\theta}{\pi} \\
~~&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=1}^{\infty} \sum_{k=0}^{\infty} a_j j r^{j-1} e^{i(j-1)\theta} \overline{b_k} r^k e^{-ik\theta} r \, dr \, \frac{d\theta}{\pi} \\~~
~~&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=1}^{\infty} \sum_{k=0}^{\infty} a_j \overline{b_k} j r^{j+k} e^{i(j-k-1)\theta} r \, dr \, \frac{d\theta}{\pi}.~~
\end{align*}
\begin{equation*}
\int_{0}^{2\pi} e^{i(j-k-1)\theta} \frac{d\theta}{\pi} = \begin{cases}
~~1 & \text{if } j = k+1, \\~~
0 & \text{if } j \neq k+1.
\end{cases}
\end{equation*}
Hence
\begin{align*}
~~\langle \frac{\partial f}{\partial z}, g angle &= \int_{0}^{\infty} e^{-r^2} \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) r^{2k+3} r \, dr \\~~
&= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
~~The other scalar product is~~
\begin{align*}
\langle f, zg angle &= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \left(\sum_{j=0}^{\infty} a_j (re^{i\theta})^j ight) \overline{\left(\sum_{k=0}^{\infty} b_k r^{k+1} e^{i(k+1)\theta} ight)} r \, dr \, \frac{d\theta}{\pi} \\
~~&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} a_j \overline{b_k} r^{j+k+1} e^{i(j-k-1)\theta} r \, dr \, \frac{d\theta}{\pi}.~~
\end{align*}
~~Again, integrating over $ \theta $ and j=k+1~~
\begin{align*}
~~\langle f, zg angle &= \int_{0}^{\infty} e^{-r^2} \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} r^{2k+3} r \, dr \\~~
&= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
\begin{align*}
~~\langle \frac{\partial f}{\partial z}, g angle &= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr, \\~~
\langle f, zg \rangle &= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
~~These are not equal due to the factor (k+1) present in $\langle \frac{\partial f}{\partial z}, g angle $. So i don't know where is the fail.~~

I'm working through this academic paper: *Stability of the Solutions of Differential Equations* whose author is Bernard Beauzamy. [A link to paper](https://typeset.io/pdf/stability-of-the-solutions-of-differential-equations-2byjzvdtk5.pdf).
In the academic paper, it works with the norm
\begin{equation*}
\left\Vert f\right\Vert= \left( \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r^2} |f(re^{i\theta})|^2 r dr\frac{d\theta}{\pi} \right)^{1/2}
\end{equation*}
for analytic polynomials
and the scalar product associated is
\begin{equation*}
\langle f,g\rangle= \int_{0}^{\infty}\int_{0}^{2\pi} e^{-r^2}f(re^{i\theta})\overline{g(re^{i\theta})} r dr\frac{d\theta}{\pi}
\end{equation*}
The lemma that I'm trying to check says the following:
For any $f$, $g$ in $\mathcal{P}_2$
\begin{equation*}
\left\langle \frac{\partial f}{\partial z} , g ight\rangle = \langle f, zg angle \qquad \langle zf,g angle = \left\langle f, \frac{\partial g}{\partial z} ight\rangle
\end{equation*}
the functions $f$ and $g$ are given by
\begin{equation*}
f(z) = \sum_{j=0}^{\infty} a_j z^j, \quad g(z) = \sum_{j=0}^{\infty} b_j z^j.
\end{equation*}
The derivative of $f$ with respect to $z$ is
\begin{equation*}
\frac{\partial f}{\partial z} = \sum_{j=1}^{\infty} a_j j z^{j-1}.
\end{equation*}
so the scalar product is
\begin{align*}
\left\langle \frac{\partial f}{\partial z}, g ight\rangle &= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \left(\sum_{j=1}^{\infty} a_j j (re^{i\theta})^{j-1} ight) \overline{\left(\sum_{k=0}^{\infty} b_k (re^{i\theta})^k ight)} r \\, dr \\, \frac{d\theta}{\pi} \\\\
&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=1}^{\infty} \sum_{k=0}^{\infty} a_j j r^{j-1} e^{i(j-1)\theta} \overline{b_k} r^k e^{-ik\theta} r \\, dr \\, \frac{d\theta}{\pi} \\\\
&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=1}^{\infty} \sum_{k=0}^{\infty} a_j \overline{b_k} j r^{j+k} e^{i(j-k-1)\theta} r \\, dr \\, \frac{d\theta}{\pi}.
\end{align*}
\begin{equation*}
\int_{0}^{2\pi} e^{i(j-k-1)\theta} \frac{d\theta}{\pi} = \begin{cases}
1 & \text{if } j = k+1, \\\\
0 & \text{if } j \neq k+1.
\end{cases}
\end{equation*}
Hence
\begin{align*}
\left\langle \frac{\partial f}{\partial z}, g ight\rangle &= \int_{0}^{\infty} e^{-r^2} \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) r^{2k+3} r \\, dr \\\\
&= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
The other scalar product is
\begin{align*}
\langle f, zg angle &= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \left(\sum_{j=0}^{\infty} a_j (re^{i\theta})^j ight) \overline{\left(\sum_{k=0}^{\infty} b_k r^{k+1} e^{i(k+1)\theta} ight)} r \\, dr \\, \frac{d\theta}{\pi} \\\\
&= \int_{0}^{\infty} \int_{0}^{2\pi} e^{-r^2} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} a_j \overline{b_k} r^{j+k+1} e^{i(j-k-1)\theta} r \\, dr \\, \frac{d\theta}{\pi}.
\end{align*}
Again, integrating over $\theta$ and $j=k+1$
\begin{align*}
\langle f, zg angle &= \int_{0}^{\infty} e^{-r^2} \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} r^{2k+3} r \\, dr \\\\
&= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
\begin{align*}
\langle \frac{\partial f}{\partial z}, g angle &= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} (k+1) \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr, \\\\
\langle f, zg \rangle &= \sum_{k=0}^{\infty} a_{k+1} \overline{b_k} \int_{0}^{\infty} e^{-r^2} r^{2k+3} dr.
\end{align*}
These are not equal due to the factor $(k+1)$ present in $\left\langle \frac{\partial f}{\partial z}, g ight\rangle$. So I don't know where the error is.

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