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Tulisan ini menjelaskan example 2, pada journal ''A Bieberbach Condition for
a Class of Pseudo-analytic Functions'' oleh J.L. Schiff and W.J.Walker. 1990.
$e^{i\theta }=r\cos \theta +ir\sin \theta =x+yi,$ $x=rcos\theta ,$ $y=r\sin \theta ,$ $r=\sqrt{x^2+y^2},$ $\theta =\arctan
y/x$ maka $\theta _x=\frac{-y}{r^2},$ $r_x=\frac xr,$ $\theta _y=\frac x{r^2},$
dan $r=\frac yr.$

Selanjutnya perhatikan operator turunan parsial berikut,

$$\frac \partial {\partial x}=\frac \partial {\partial r}\frac{\partial r}{%
\partial x}+\frac \partial {\partial \theta }\frac{\partial \theta }{%
\partial x}=\frac xr\frac \partial {\partial r}-\frac y{r^2}\frac \partial
{\partial \theta }$$

$$\frac \partial {\partial y}=\frac \partial {\partial r}\frac{\partial r}{%
\partial y}+\frac \partial {\partial \theta }\frac{\partial \theta }{%
\partial y}=\frac yr\frac \partial {\partial r}+\frac x{r^2}\frac \partial
{\partial \theta },$$ dan operator L akan menjadi,

\begin{eqnarray*}
L &=&\frac \partial {\partial x}+i\frac \partial {\partial y} \\
&=&\frac xr\frac \partial {\partial r}-\frac y{r^2}\frac \partial {\partial
\theta }+i\left( \frac yr\frac \partial {\partial r}+\frac x{r^2}\frac
\partial {\partial \theta }\right) \\
&=&\left( \frac xr+\frac yri\right) \frac \partial {\partial r}+\left( \frac{%
-y}{r^2}+\frac x{r^2}i\right) \frac \partial {\partial \theta } \\
&=&\frac 1r\left[ \left( x+yi\right) \frac \partial {\partial r}+\frac
1r\left( -y+xi\right) \frac \partial {\partial \theta }\right] \\
&=&\frac 1r\left[ \left( r\cos \theta +ri\sin \theta \right) \frac \partial
{\partial r}+\frac 1r\left( r\cos \theta +ir\sin \theta \right) i\frac
\partial {\partial \theta }\right] \\
&=&\left( \cos \theta +i\sin \theta \right) \frac \partial {\partial
r}+\frac 1r\left( \cos \theta +i\sin \theta \right) i\frac \partial
{\partial \theta } \\
&=&\left( \cos \theta +\sin \theta i\right) (\frac \partial {\partial
r}+\frac 1ri\frac \partial {\partial \theta }) \\
&=&e^{i\theta }(\frac \partial {\partial r}+\frac 1ri\frac \partial
{\partial \theta })
\end{eqnarray*}
atau bisa juga dengan memandang $x=\frac{z+\overline{z}}2$ dan $y=\frac{z-%
\overline{z}}{2i}$ maka \begin{eqnarray*}
\frac \partial {\partial \overline{z}} &=&\frac \partial {\partial x}\frac{%
\partial x}{\partial \overline{z}}+\frac \partial {\partial y}\frac{\partial
y}{\partial \overline{z}} \\
&=&\frac 12\frac \partial {\partial x}-\frac 1{2i}\frac \partial {\partial y}
\\
&=&\frac 12\left( \frac \partial {\partial x}+i\frac \partial {\partial
y}\right)
\end{eqnarray*}
\[
L=\frac \partial {\partial x}+i\frac \partial {\partial y}=2\frac \partial
{\partial \overline{z}} 
\]

\[
\Psi _n(\mu r)=1+\frac{\left( \frac{\mu r}2\right) ^2}{1.(n+1)}+\frac{\left( 
\frac{\mu r}2\right) ^4}{1.2.(n+1)(n+2)}+\frac{\left( \frac{\mu r}2\right) ^6%
}{1.2.3.(n+1)(n+2)(n+3)}+... 
\]

\begin{eqnarray*}
\frac \partial {\partial r}(\Psi _n(\mu r)) &=&\frac{2\left( \frac{\mu r}%
2\right) \frac \mu 2}{1.(n+1)}+\frac{4\left( \frac{\mu r}2\right) ^3\frac
\mu 2}{1.2.(n+1)(n+2)}+\frac{6\left( \frac{\mu r}2\right) ^5\frac \mu 2}{%
1.2.3.(n+1)(n+2)(n+3)}+... \\
&=&\frac{\frac{\mu r}2}{n+1}\left\{ \frac{2.\frac \mu 2}1+\frac{4\left( 
\frac{\mu r}2\right) ^2\frac \mu 2}{1.2.(n+2)}+\frac{6\left( \frac{\mu r}%
2\right) ^4\frac \mu 2}{1.2.3.(n+2)(n+3)}+...\right\} \\
&=&\frac{\frac{\mu r}2}{n+1}\left\{ 1+\frac{\left( \frac{\mu r}2\right)
^2\mu }{1.(n+2)}+\frac{\left( \frac{\mu r}2\right) ^4\mu }{1.2.(n+2)(n+3)}+%
\frac{\left( \frac{\mu r}2\right) ^6\mu }{1.2.3.(n+2)(n+3)(n+4)}+..\right\}
\\
&=&\frac{\frac{\mu ^2r}2}{n+1}\left\{ 1+\frac{\left( \frac{\mu r}2\right) ^2%
}{1.(n+2)}+\frac{\left( \frac{\mu r}2\right) ^4}{1.2.(n+2)(n+3)}+\frac{%
\left( \frac{\mu r}2\right) ^6}{1.2.3.(n+2)(n+3)(n+4)}+..\right\} \\
&=&\frac{\mu ^2r}{2(n+1)}\Psi _{n+1}(\mu r)
\end{eqnarray*}

\[
\Psi _{n+1}(\mu r)=1+\frac{\left( \frac{\mu r}2\right) ^2}{1.(n+2)}+\frac{%
\left( \frac{\mu r}2\right) ^4}{1.2.(n+2)(n+3)}+\frac{\left( \frac{\mu r}%
2\right) ^6}{1.2.3.(n+2)(n+3)(n+4)}+... 
\]

\begin{eqnarray*}
\frac \partial {\partial r}(\Psi _{n+1}(\mu r)) &=&\frac{2\left( \frac{\mu r}%
2\right) \frac \mu 2}{1.(n+2)}+\frac{4\left( \frac{\mu r}2\right) ^3\frac
\mu 2}{1.2.(n+2)(n+3)}+\frac{6\left( \frac{\mu r}2\right) ^5\frac \mu 2}{%
1.2.3.(n+2)(n+3)(n+4)}+... \\
&=&\frac{\frac{\mu r}2}{n+2}\left\{ \frac{2.\frac \mu 2}1+\frac{4\left( 
\frac{\mu r}2\right) ^2\frac \mu 2}{1.2.(n+3)}+\frac{6\left( \frac{\mu r}%
2\right) ^4\frac \mu 2}{1.2.3.(n+3)(n+4)}+...\right\} \\
&=&\frac{\frac{\mu r}2\mu }{n+2}\left\{ 1+\frac{\left( \frac{\mu r}2\right)
^2\mu }{1.(n+3)}+\frac{\left( \frac{\mu r}2\right) ^4\mu }{1.2.(n+3)(n+4)}+%
\frac{\left( \frac{\mu r}2\right) ^6\mu }{1.2.3.(n+3)(n+4)(n+5)}+..\right\}
\\
&=&\frac{\frac{\mu ^2r}2}{n+2}\left\{ 1+\frac{\left( \frac{\mu r}2\right) ^2%
}{1.(n+3)}+\frac{\left( \frac{\mu r}2\right) ^4}{1.2.(n+3)(n+4)}+\frac{%
\left( \frac{\mu r}2\right) ^6}{1.2.3.(n+3)(n+4)(n+5)}+..\right\} \\
&=&\frac{\mu ^2r}{2(n+2)}\Psi _{n+2}(\mu r)
\end{eqnarray*}

Secara umum dapat diperlihatkan bahwa, untuk sembarang k bilangan bulat tak
negatif, berlaku 
\[
\frac \partial {\partial r}(\Psi _{n+k}(\mu r))=\frac{\mu ^2r}{2(n+k)}\Psi
_{n+k+1}(\mu r) 
\]

\begin{example}
Following Duffin[5, p.124], we take $u=ce^{in\theta }I_n(\mu r),$ where $%
c=n!2^n\mu ^{-n-1}.$ If we write 
\[
I_n(x)=\frac 1{n!}\left( \frac x2\right) ^n\Psi _n(x)
\]
\end{example}

where 
\[
\Psi _n(x)=1+\frac{\left( \frac x2\right) ^2}{1.(n+1)}+\frac{\left( \frac
x2\right) ^4}{1.2.(n+1)(n+2)}+\frac{\left( \frac x2\right) ^6}{%
1.2.3.(n+1)(n+2)(n+3)}+..., 
\]

we have 
\begin{eqnarray*}
u &=&n!2^n\mu ^{-n-1}e^{in\theta }I_n(\mu r) \\
&=&n!2^n\mu ^{-n-1}e^{in\theta }\frac 1{n!}\left( \frac{\mu r}2\right)
^n\Psi _n(\mu r) \\
&=&\mu ^{-n-1}e^{in\theta }\left( \mu r\right) ^n\Psi _n(\mu r) \\
&=&\frac 1\mu e^{in\theta }r^n\Psi _n(\mu r) \\
&=&\frac 1\mu (e^{i\theta }r)^n\Psi _n(\mu r) \\
&=&\frac 1\mu (z)^n\Psi _n(\mu r),\text{ }
\end{eqnarray*}

where $z=re^{i\theta }.$ Note that 
\[
L=\frac \partial {\partial x}+i\frac \partial {\partial y}=e^{i\theta
}\left( \frac \partial {\partial r}+\frac 1ri\frac \partial {\partial \theta
}\right) 
\]

and since also $L=2\frac \partial {\partial \overline{z}}\stackrel{}{}$it
follows that
\begin{eqnarray*}
L(z^n\psi _n(\mu r)) &=&L((e^{i\theta }r)^n\psi _n(\mu r)) \\
&=&e^{i\theta }.n(e^{i\theta }r)^{n-1}.e^{i\theta }\psi _n(\mu r)+e^{i\theta
}(e^{i\theta }r)^n(\psi _n(\mu r))^{\prime }+e^{i\theta }.\frac
irnire^{i\theta }(e^{i\theta }r)^{n-1}.\psi _n(\mu r) \\
&=&e^{i\theta }(e^{i\theta }r)^n(\psi _n(\mu r))^{\prime } \\
&=&z^ne^{i\theta }(\psi _n(\mu r))^{\prime } \\
&=&z^ne^{i\theta }\frac{\mu ^2r}{2(n+1)}\psi _{n+1}(\mu r) \\
&=&\frac{\mu ^2z^{n+1}}{2(n+1)}\psi _{n+1}(\mu r)
\end{eqnarray*}

Hence a $\mu $-regular function called the nth pseudo-power is defined by
\begin{eqnarray*}
z^{(n)} &=&\mu u+\overline{Lu} \\
&=&\mu z^n\frac{\psi _n(\mu r)}\mu +\overline{L(\frac{z^n\psi _n(\mu r)}\mu )%
} \\
&=&z^n\psi _n(\mu r)+\frac \mu {2(n+1)}\overline{z^{n+1}}\psi _{n+1}(\mu r)
\end{eqnarray*}

\begin{theorem}

Misal f(z) $\mu -regular$ pada cakram $\left| z\right| \leq a.$

Maka untuk $\left| z\right| <a,$%
\[
f(z)=\sum_{n=0}^\infty a_nz^n\psi _n(\mu r)+\sum_{n=0}^\infty \overline{a_n}%
\frac \mu {2(n+1)}\overline{z^{n+1}}\psi _{n+1}(\mu r),
\]

dimana

\[
a_n=\frac 1{2\pi a^n\psi _n(\mu a)}\int_0^{2\pi }e^{-in\theta }f(ae^{i\theta
})d\theta 
\]
\end{theorem}

Bukti

\end{document}
