# Complex Analysis/Exercises/Sheet 2/Exercise 3
#### Task (Working with Polynomials, 5 Points)

Consider a polynomial $p\colon \mathbf {C} \to \mathbf {C}$, given by
$p(z)=\sum _{0\leq \kappa \leq k \atop 0\leq \lambda \leq \ell }a_{\kappa \lambda }x^{\kappa }y^{\lambda }$
with $x=\mathop {\rm {Re}} z$ and $y=\mathop {\rm {Im}} z$. Show that $p$ can also be expressed as a polynomial in $z$
and ${\bar {z}}$ by specifying the coefficients in
$p(z)=\sum _{0\leq \mu \leq m \atop 0\leq \nu \leq n}b_{\mu \nu }z^{\mu }{\bar {z}}^{\nu }$

#### Solution

For $z=x+iy$,is ${\bar {z}}=x-iy$. Solving the system
${\begin{array}{rl}z&=x+iy\\{\bar {z}}&=x-iy\end{array}}$
for $x,y$, we get $x={\frac {1}{2}}(z+{\bar {z}})$ and $y={\frac {1}{2i}}(z-{\bar {z}})$. We substitute these into $p$. Using the Binomial theorem, we get
${\begin{array}{rl}p(z)&=\displaystyle \sum _{\kappa ,\lambda }a_{\kappa \lambda }x^{\kappa }y^{\lambda }\\&=\displaystyle \sum _{\kappa ,\lambda }a_{\kappa \lambda }{\frac {1}{2^{\kappa }}}(z+{\bar {z}})^{\kappa }{\frac {1}{(2i)^{\lambda }}}(z-{\bar {z}})^{\lambda }\\&=\displaystyle \sum _{\kappa ,\lambda }{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}\sum _{\alpha =0}^{\kappa }{\binom {\kappa }{\alpha }}z^{\alpha }{\bar {z}}^{\kappa -\alpha }\cdot \sum _{\beta =0}^{\lambda }(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\beta }{\bar {z}}^{\lambda -\beta }\\&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\alpha =0}^{\kappa }\sum _{\beta =0}^{\lambda }{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\alpha }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\alpha +\beta }{\bar {z}}^{\kappa +\lambda -(\alpha +\beta )}\end{array}}$
We then perform an index transformation, replacing the summation over $\alpha$ with a summation over $\mu :=\alpha +\beta$. Since $0\leq \alpha \leq \kappa ,0\leq \beta \leq \lambda$,
$\mu$ ranges from $0$ to $\kappa +\lambda$. For a fixed $\mu$, we consider $\beta$ values between $0$ and $\lambda$, where
$0\leq \mu -\beta \leq \kappa$, so $\mu -\kappa \leq \beta \leq \mu$. Hence, $\beta$ ranges from $\max(0,\mu -\kappa )$ to $\min(\mu ,\lambda )$. We get
${\begin{array}{rl}p(z)&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\alpha =0}^{\kappa }\sum _{\beta =0}^{\lambda }{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\alpha }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\alpha +\beta }{\bar {z}}^{\kappa +\lambda -(\alpha +\beta )}\\&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\mu =0}^{\kappa +\lambda }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\end{array}}$
We then switch the order of summation. We have $0\leq \mu \leq k+\ell$ and for a fixed $\mu$, $\kappa +\lambda \geq \mu$, which means $\lambda \geq \mu -\kappa$,
resulting in
${\begin{array}{rl}p(z)&=\displaystyle \sum _{\kappa ,\lambda }\sum _{\mu =0}^{\kappa +\lambda }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\\&=\displaystyle \sum _{\mu =0}^{k+\ell }\sum _{\kappa =0}^{\mu }\sum _{\lambda =\mu -\kappa }^{\ell }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\end{array}}$
Finally, we perform another index transformation. We replace $\kappa$ with $\nu :=\kappa +\lambda -\mu$. $\nu$ ranges from $0$ to $k+\ell$. For a fixed $\nu$,
we have $\kappa =\nu +\mu -\lambda$. Therefore, we consider only values of $\lambda$ for which $0\leq \nu +\mu -\lambda \leq \mu$, i.e., $\nu \leq \lambda \leq \nu +\mu$.
We get
${\begin{array}{rl}p(z)&=\displaystyle \sum _{\mu =0}^{k+\ell }\sum _{\kappa =0}^{\mu }\sum _{\lambda =\mu -\kappa }^{\ell }\sum _{\beta =\max(0,\mu -\kappa )}^{\min(\mu ,\lambda )}{\frac {a_{\kappa \lambda }}{2^{\kappa +\lambda }i^{\lambda }}}{\binom {\kappa }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}z^{\mu }{\bar {z}}^{\kappa +\lambda -\mu }\\&=\displaystyle \sum _{\mu =0}^{k+\ell }\sum _{\nu =0}^{k+\ell }\left(\sum _{\lambda =\nu }^{\min(\ell ,\nu +\mu )}\sum _{\beta =\lambda -\nu }^{\min(\mu ,\lambda )}{\frac {a_{\mu +\nu -\lambda ,\lambda }}{2^{\mu +\nu }i^{\lambda }}}{\binom {\mu +\nu -\lambda }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }}\right)z^{\mu }{\bar {z}}^{\nu }\end{array}}$
Thus,
$b_{\mu \nu }=\sum _{\lambda =\nu }^{\min(\ell ,\nu +\mu )}\sum _{\beta =\lambda -\nu }^{\min(\mu ,\lambda )}{\frac {a_{\mu +\nu -\lambda ,\lambda }}{2^{\mu +\nu }i^{\lambda }}}{\binom {\mu +\nu -\lambda }{\mu -\beta }}(-1)^{\lambda -\beta }{\binom {\lambda }{\beta }},\qquad 0\leq \mu ,\nu \leq k+\ell$

### Translation and Version Control

This page was translated based on the following Wikiversity source page and uses the concept of Translation and Version Control for a transparent language fork in a Wikiversity:
* Source: Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_3 - URL:
https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_3
* Date: 01/14/2024