Kurs:Funktionentheorie/Übungen/2. Zettel/Aufgabe 3

Wir betrachten ein Polynom ${\displaystyle p\colon \mathbf {C} \to \mathbf {C} }$, gegeben durch

${\displaystyle p(z)=\sum _{0\leq \kappa \leq k \atop 0\leq \lambda \leq \ell }a_{\kappa \lambda }x^{\kappa }y^{\lambda }}$

mit ${\displaystyle x=\mathop {\rm {Re}} z}$ und ${\displaystyle y=\mathop {\rm {Im}} z}$. Zeige, dass sich ${\displaystyle p}$ auch als Polynom in ${\displaystyle z}$ und ${\displaystyle {\bar {z}}}$ darstellen lässt, indem Du die Koeffizienten in

${\displaystyle p(z)=\sum _{0\leq \mu \leq m \atop 0\leq \nu \leq n}b_{\mu \nu }z^{\mu }{\bar {z}}^{\nu }}$

angibst.

Für ${\displaystyle z=x+iy}$ ist ${\displaystyle {\bar {z}}=x-iy}$. Lösen wir das Gleichungssystem

${\displaystyle {\begin{array}{rl}z&=x+iy\\{\bar {z}}&=x-iy\end{array}}}$

nach ${\displaystyle x,y}$ auf, erhalten wir ${\displaystyle x={\frac {1}{2}}(z+{\bar {z}})}$ und ${\displaystyle y={\frac {1}{2i}}(z-{\bar {z}})}$. Dies setzen wir in ${\displaystyle p}$ ein, mit dem Binomialtheorem folgt

${\displaystyle {\begin{array}{rl}p(z)&=\displaystyle \sum _{\kappa ,\lambda }a_{\kappa \lambda }x^{\kappa }y^{\lambda }\\&=\displaystyle \sum _{\kappa ,\lambda }a_{\kappa \lambda }{\frac {1}{2^{\kappa }}}(z+{\bar {z}})^{\kappa }{\frac {1}{(2i)^{\lambda }}}(z-{\bar {z}})^{\lambda }\\&=\displaystyle \sum _{\kappa ,\lambda }{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}\sum _{\alpha =0}^{\kappa }{\binom {\kappa }{\alpha }}z^{\alpha }{\bar {z}}^{\kappa -\alpha }\cdot \sum _{\beta =0}^{\lambda }(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\beta }{\bar {z}}^{\lambda -\beta }\\&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\alpha =0}^{\kappa }\sum _{\beta =0}^{\lambda }{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\alpha }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\alpha +\beta }{\bar {z}}^{\kappa +\lambda -(\alpha +\beta )}\end{array}}}$

Nun führen wir eine Indextransformation durch, wir ersetzen die Summation über ${\displaystyle \alpha }$ durch eine Summation über ${\displaystyle \mu :=\alpha +\beta }$. Wegen ${\displaystyle 0\leq \alpha \leq \kappa ,0\leq \beta \leq \lambda }$ läuft ${\displaystyle \mu }$ von ${\displaystyle 0}$ bis ${\displaystyle \kappa +\lambda }$. Für festes ${\displaystyle \mu }$ müssen wir für ${\displaystyle \beta }$ die Werte zwischen ${\displaystyle 0}$ und ${\displaystyle \lambda }$ betrachten, für die ${\displaystyle 0\leq \mu -\beta \leq \kappa }$, also ${\displaystyle \mu -\kappa \leq \beta \leq \mu }$ ist. Daher läuft ${\displaystyle \beta }$ von ${\displaystyle \max(0,\mu -\kappa )}$ bis ${\displaystyle \min(\mu ,\lambda )}$. Wir erhalten

${\displaystyle {\begin{array}{rl}p(z)&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\alpha =0}^{\kappa }\sum _{\beta =0}^{\lambda }{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\alpha }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\alpha +\beta }{\bar {z}}^{\kappa +\lambda -(\alpha +\beta )}\\&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\mu =0}^{\kappa +\lambda }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\end{array}}}$

Nun tauschen wir die Summationsreihenfolge, es ist ${\displaystyle 0\leq \mu \leq k+\ell }$ und für festes ${\displaystyle \mu }$ muss stets ${\displaystyle \kappa +\lambda \geq \mu }$ gelten, d. h. ${\displaystyle \lambda \geq \mu -\kappa }$, wir erhalten

${\displaystyle {\begin{array}{rl}p(z)&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\mu =0}^{\kappa +\lambda }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\\&=\displaystyle \sum _{\mu =0}^{k+\ell }\sum _{\kappa =0}^{\mu }\sum _{\lambda =\mu -\kappa }^{\ell }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\end{array}}}$

Nun noch eine letzte Indextransformation. Wir ersetzen ${\displaystyle \kappa }$ durch ${\displaystyle \nu :=\kappa +\lambda -\mu }$. ${\displaystyle \nu }$ läuft von ${\displaystyle 0}$ bis ${\displaystyle k+\ell }$. Für festes ${\displaystyle \nu }$ ist ${\displaystyle \kappa =\nu +\mu -\lambda }$. Es sind also nur die Werte von ${\displaystyle \lambda }$ zulässig, für die ${\displaystyle 0\leq \nu +\mu -\lambda \leq \mu }$ ist, d. h. ${\displaystyle \nu \leq \lambda \leq \nu +\mu }$ gilt. Wir erhalten

${\displaystyle {\begin{array}{rl}p(z)&=\displaystyle \sum _{\mu =0}^{k+\ell }\sum _{\kappa =0}^{\mu }\sum _{\lambda =\mu -\kappa }^{\ell }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\\&=\displaystyle \sum _{\mu =0}^{k+\ell }\sum _{\nu =0}^{k+\ell }\left(\sum _{\lambda =\nu }^{\min(\ell ,\nu +\mu )}\sum _{\beta =\lambda -\nu }^{\min(\mu ,\lambda )}{\frac {a_{\mu +\nu -\lambda ,\lambda }}{2^{\mu +\nu }i^{\lambda }}}{\binom {\mu +\nu -\lambda }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}\right)z^{\mu }{\bar {z}}^{\nu }\end{array}}}$

Also ist

${\displaystyle b_{\mu \nu }=\sum _{\lambda =\nu }^{\min(\ell ,\nu +\mu )}\sum _{\beta =\lambda -\nu }^{\min(\mu ,\lambda )}{\frac {a_{\mu +\nu -\lambda ,\lambda }}{2^{\mu +\nu }i^{\lambda }}}{\binom {\mu +\nu -\lambda }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }},\qquad 0\leq \mu ,\nu \leq k+\ell }$