Complex Analysis/Exercises/Sheet 2/Exercise 3

Consider a polynomial ${\displaystyle p\colon \mathbf {C} \to \mathbf {C} }$, given by

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

with ${\displaystyle x=\mathop {\rm {Re}} z}$ and ${\displaystyle y=\mathop {\rm {Im}} z}$. Show that ${\displaystyle p}$ can also be expressed as a polynomial in ${\displaystyle z}$ and ${\displaystyle {\bar {z}}}$ by specifying the coefficients in

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

.

For ${\displaystyle z=x+iy}$,is ${\displaystyle {\bar {z}}=x-iy}$. Solving the system

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

for ${\displaystyle x,y}$, we get ${\displaystyle x={\frac {1}{2}}(z+{\bar {z}})}$ and ${\displaystyle y={\frac {1}{2i}}(z-{\bar {z}})}$. We substitute these into ${\displaystyle p}$. Using the Binomial theorem, we get

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

We then perform an index transformation, replacing the summation over ${\displaystyle \alpha }$ with a summation over ${\displaystyle \mu$ :=\alpha +\beta } . Since ${\displaystyle 0\leq \alpha \leq \kappa ,0\leq \beta \leq \lambda }$, ${\displaystyle \mu }$ ranges from ${\displaystyle 0}$ to ${\displaystyle \kappa +\lambda }$. For a fixed ${\displaystyle \mu }$, we consider ${\displaystyle \beta }$ values between ${\displaystyle 0}$ and ${\displaystyle \lambda }$, where ${\displaystyle 0\leq \mu -\beta \leq \kappa }$, so ${\displaystyle \mu -\kappa \leq \beta \leq \mu }$. Hence, ${\displaystyle \beta }$ ranges from ${\displaystyle \max(0,\mu -\kappa )}$ to ${\displaystyle \min(\mu ,\lambda )}$. We get

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

We then switch the order of summation. We have ${\displaystyle 0\leq \mu \leq k+\ell }$ and for a fixed ${\displaystyle \mu }$, ${\displaystyle \kappa +\lambda \geq \mu }$, which means ${\displaystyle \lambda \geq \mu -\kappa }$, resulting in

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

Finally, we perform another index transformation. We replace ${\displaystyle \kappa }$ with ${\displaystyle \nu$ :=\kappa +\lambda -\mu } . ${\displaystyle \nu }$ ranges from ${\displaystyle 0}$ to ${\displaystyle k+\ell }$. For a fixed ${\displaystyle \nu }$, we have ${\displaystyle \kappa =\nu +\mu -\lambda }$. Therefore, we consider only values of ${\displaystyle \lambda }$ for which ${\displaystyle 0\leq \nu +\mu -\lambda \leq \mu }$, i.e., ${\displaystyle \nu \leq \lambda \leq \nu +\mu }$. We get

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

Thus,

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

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• Source: Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_3 - URL:

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• Date: 01/14/2024