Complex Analysis/Exercises/Sheet 2/Task 3

Task (Working with Polynomials, 5 Points)

Consider a polynomial $p:𝐂\to 𝐂$, given by

$p(z)=\sum _{\genfrac{}{}{0ex}{}{0\le \kappa \le k}{0\le \lambda \le \ell }}{a}_{\kappa \lambda }{x}^{\kappa }{y}^{\lambda }$

with $x=\mathrm{R}\mathrm{e}z$ and $y=\mathrm{I}\mathrm{m}z$. Show that $p$ can also be expressed as a polynomial in $z$ and $\overline{z}$ by specifying the coefficients in

$p(z)=\sum _{\genfrac{}{}{0ex}{}{0\le \mu \le m}{0\le \nu \le n}}{b}_{\mu \nu }{z}^{\mu }{\overline{z}}^{\nu }$

.

Solution

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

$\begin{array}{cc}z& =x+iy\\ \overline{z}& =x-iy\end{array}$

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

$\begin{array}{cc}p(z)& =\sum _{\kappa ,\lambda }{a}_{\kappa \lambda }{x}^{\kappa }{y}^{\lambda }\\ & =\sum _{\kappa ,\lambda }{a}_{\kappa \lambda }\frac{1}{2^{\kappa }}(z+\overline{z}{)}^{\kappa }\frac{1}{(2i{)}^{\lambda }}(z-\overline{z}{)}^{\lambda }\\ & =\sum _{\kappa ,\lambda }\frac{a_{\kappa \lambda }}{2^{\kappa +\lambda }{i}^{\lambda }}\sum _{\alpha =0}^{\kappa }(\genfrac{}{}{0ex}{}{\kappa }{\alpha })z^{\alpha }{\overline{z}}^{\kappa -\alpha }\cdot \sum _{\beta =0}^{\lambda }(-1{)}^{\lambda -\beta }(\genfrac{}{}{0ex}{}{\lambda }{\beta })z^{\beta }{\overline{z}}^{\lambda -\beta }\\ & =\sum _{\kappa ,\lambda }\sum _{\alpha =0}^{\kappa }\sum _{\beta =0}^{\lambda }\frac{a_{\kappa \lambda }}{2^{\kappa +\lambda }{i}^{\lambda }}(\genfrac{}{}{0ex}{}{\kappa }{\alpha })(-1{)}^{\lambda -\beta }(\genfrac{}{}{0ex}{}{\lambda }{\beta })z^{\alpha +\beta }{\overline{z}}^{\kappa +\lambda -(\alpha +\beta )}\end{array}$

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

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

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

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

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 $0\le \nu +\mu -\lambda \le \mu$, i.e., $\nu \le \lambda \le \nu +\mu$. We get

$\begin{array}{cc}p(z)& =\sum _{\mu =0}^{k+\ell }\sum _{\kappa =0}^{\mu }\sum _{\lambda =\mu -\kappa }^{\ell }\sum _{\beta =\max (0,\mu -\kappa )}^{\min (\mu ,\lambda )}\frac{a_{\kappa \lambda }}{2^{\kappa +\lambda }{i}^{\lambda }}(\genfrac{}{}{0ex}{}{\kappa }{\mu -\beta })(-1{)}^{\lambda -\beta }(\genfrac{}{}{0ex}{}{\lambda }{\beta })z^{\mu }{\overline{z}}^{\kappa +\lambda -\mu }\\ & =\sum _{\mu =0}^{k+\ell }\sum _{\nu =0}^{k+\ell }(\sum _{\lambda =\nu }^{\min (\ell ,\nu +\mu )}\sum _{\beta =\lambda -\nu }^{\min (\mu ,\lambda )}\frac{a_{\mu +\nu -\lambda ,\lambda }}{2^{\mu +\nu }{i}^{\lambda }}(\genfrac{}{}{0ex}{}{\mu +\nu -\lambda }{\mu -\beta })(-1{)}^{\lambda -\beta }(\genfrac{}{}{0ex}{}{\lambda }{\beta }))z^{\mu }{\overline{z}}^{\nu }\end{array}$

Thus,

$b_{\mu \nu }=\sum _{\lambda =\nu }^{\min (\ell ,\nu +\mu )}\sum _{\beta =\lambda -\nu }^{\min (\mu ,\lambda )}\frac{a_{\mu +\nu -\lambda ,\lambda }}{2^{\mu +\nu }{i}^{\lambda }}(\genfrac{}{}{0ex}{}{\mu +\nu -\lambda }{\mu -\beta })(-1{)}^{\lambda -\beta }(\genfrac{}{}{0ex}{}{\lambda }{\beta }),0\le \mu ,\nu \le k+\ell$

Translation and Version Control

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• Source: Kurs:Funktionentheorie/Exercises/Sheet 2/Task 3 - URL:

https://de.wikiversity.org/wiki/Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_3

• Date: 01/14/2024

de:Kurs:Funktionentheorie/Übungen/2._Zettel/Aufgabe_3