Exercise 1.1.3 (Conjugacy)

Question

Let $ z, z' \in \mathbb{C}$. Then $ z$ and $ z' $ are conjugate over $\mathbb{R}$ if and only if $z'=z$ or $z'= \bar z $. Both proofs of ‘if’ contain little gaps: ‘It follows by
induction’ in the first proof, and ‘it’s easy to see’ in the second. Fill them.

Hassaan Naeem · Accepted Answer

\paragraph{Solution:} We show both both parts (i) and (ii) seperately

\begin{enumerate}
    \item[(i)] Follows from induction that for any polynomial $p$ over $\mathbb{R}$, $\overline{ p(w)} = p(\overline w)$:\
        \linebreak 
        Let $p(w) = c_0 + c_1w^1 + c_2w^2 + ... + c_nw^n$ where $w^n \in \mathbb{C}$ and $c_n \in \mathbb{C}$. \
            \begin{equation*}
                \begin{aligned}
                    \overline{p(w)} & = \overline{c_0 +c_1w^1 + c_2w^2 + ... + c_nw^n}\
                    & = \overline{c_0} + \overline{c_1w^1} + \overline{c_2w^2} + ... + \overline{c_nw^n}\
                    & = c_0 + c_1\overline{w^1} + c_2\overline{w^2} + ... + c_n\overline{w^n}\
                    & = p(\overline w)
                \end{aligned}
            \end{equation*}
    \item[(ii)] Checking that r is the zero polynomial:\
        \linebreak
        	extbf{Lemma}. If $r(x) = a_0 + a_1x^1+ ... + a_nx^n $ and $r(x) = 0,   \forall x 
eq 0 $, then $a_0 = 0$. Since $\mathbb{Q}$ is a field, and must contain a zero.\
        
        By this lemma we have that $\forall x$, $ r(x) = 0$ and therefore\
        $r(x) = 0 = x(a_1+ a_2x + ... + a_nx^{n-1}) = a_1+ a_2x + ... + a_nx^{n-1}$ and $\forall x 
eq 0$ $a_1 = 0$.
        Hence, we repeat the Lemma and show that all $a_1, ..., a_n = 0$. Therefore $r(x)$ is the zero polynomial.
\end{enumerate}

Richard Ganaye · Answer

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\begin{proof} 
\begin{enumerate}
\item[(i)] We prove by induction that, for any polynomial $p(t) \in \R[t]$, and for any complex number $w$, $\overline{p(w)} = p(\overline{w})$. Consider the proposition, where $n \in \N =\Z_{\geq 0}$,
$$\mathscr{P}(n) : \qquad \forall p \in \R[t], \ \deg(p) \leq n \Rightarrow \forall w \in \C,\ \overline{p(w)} = p(\overline{w}).$$
$\bullet$ $\mathscr{P}(0)$ : If $\deg(p) \leq 0$, where $p \in \R[t]$, then $p = a \in \C$ is a constant complex. Thus $\overline{p(w)} = \overline{a} = p(\overline{w})$. This shows that $\mathscr{P}(0)$ is true.

$\bullet$ Assume now the induction hypothesis $\mathscr{P}(n)$: for any polynomial $p$ such that $\deg(p) \leq n$, and for any complex $w$, $\overline{p(w)} = p(\overline{w})$.

(We suppose that the induction proof of $\overline{w^n} = \overline{w}^n (w \in \C , n \in \N)$ is already done :

$\overline{w^0} = \overline{1} = 1 = \overline{w}^0$; if $\overline{w^n} = \overline{w}^n$ for some $n \in \N$, then $\overline{w^{n+1}} = \overline{w\cdot w^n} = \overline{w} \cdot \overline{w^n} =  \overline{w} \cdot \overline{w}^n = \overline{w}^{n+1}$.)

Let now $q$ be a polynomial of degree $n+1$. Then $q(t) = at^{n+1} + p(t)$, where $a \in \R, p \in \R[t]$ satisfies $\deg(p) \leq n$ ($t$ is a variable). Using the induction hypothesis on $p$, we obtain for any $w \in \C$,
\begin{align*}
\overline{q(w)} &= \overline{aw^{n+1} + p(w)}\
&=a\overline{w^{n+1}} + \overline{p(w)}\
&= a \overline{w}^{n+1} + p(\overline{w})\qquad 	ext{(induction hypothesis)}\
&= q(\overline{w}).
\end{align*}
This proves $\mathscr{P}(n+1)$, and the induction is done.

$\bullet$ To conclude, 
$$ \forall p \in \R[t], \ \forall w \in \C,\ \overline{p(w)} = p(\overline{w}).$$
Therefore, if $p\in \R[t]$, and $w \in \C$, $p(w) = 0 \iff p(\overline{w}) = 0$.

\item[(ii)] Write $z = x + iy$ with $x,y \in \R$, some fixed complex nomber. Let $p \in \R[t]$ such that $p(z) = 0$. We will prove that $p(\overline{z}) = 0$. If $y = 0$, then $p(\overline{z}) = p(x) = p(z) = 0$, so we can suppose that $y
e 0$.

Let $m(t) = (t-x)^2 + y^2 \in \R[t]$. Then $m(z) = 0$. The Euclidean division of $p(t)$ by $m(t)$ gives
$$p(t) = m(t) q(t) + r(t),\qquad q(t) \in \R[t], \ r(t) \in \R[t], \ \deg(r) < 2.$$
Thus $r(t) = at+b,\ a,b \in \R$. Since $p(z) = 0$ and $m(z) = 0$, we obtain $az +b = 0$, that is 
$$0 =a(x+ iy) + b = (ax+b) + iay.$$
Here $ax+b \in \R,\ y\in \R$. This implies $ax+b = 0$ and $ay = 0$. Since $y 
e 0$, we obtain $a= 0$, and $b = -ay = 0$. Therefore $r(t) = at + b= 0$ is the null polynomial. This gives $p(t) = m(t) q(t)$. But $m(\overline{z}) = 0$, so $p(\overline{z}) = 0$.
\end{enumerate}
\end{proof}

Exercise 1.1.3 (Conjugacy)

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Exercise 1.1.3 (Conjugacy)

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