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$$\begin{align}
\text{Question:} \\
\text{Given that:} \\
&a_{1}z^{n} + a_{2}z^{n-1} +\dots+a_{n-1}z^{1}+a_{n}=0, \\
&\text{Where }a_{k} \text{ is a real coefficient,} \\
\text{Show that:} \\
&a_{1}(\overline{z})^{n} + a_{2}(\overline{z})^{n-1} +\dots+a_{n-1}(\overline{z})^{1}+a_{n}=0
\end{align}
$$
$$\begin{align}
\text{Let }z&=r cis(\theta) \\
\therefore \overline{z}&=rcis(-\theta) \\
\text{Hence, }z^{n}&=r^{n}cis(n\theta)\text{ (De Moivre's Theorem)} \\
\text{Similarly, }(\overline{z})^{n}&=r^{n}cis(-n\theta) \\
\text{Note that if }&c=acis(b) \text{ then }\overline{c}=acis(-b) \\
\text{Hence, }&z^{n},(\overline{z})^{n}\text{ is such that } a=r^{n},b=n\theta \\
\therefore (\overline{z})^{n}&=\overline{z^{n}}
\end{align}
$$
$$\begin{align}
\text{So our polynomial can be }&\text{rewritten as such:} \\
a_{1}(\overline{z})^{n} + a_{2}(\overline{z})^{n-1} +\dots&=a_{1}\overline{z^{n}} + a_{2}\overline{z^{n-1}} +\dots \\
\text{(Simplified for space)} \\
\text{Note that for real coefficient }a, \\
a \overline{z}=\overline{az}\text{ (Property of }&\text{complex number)} \\
\therefore a_{1}\overline{z^{n}} + a_{2}\overline{z^{n-1}} +\dots&=\overline{a_{1}z^{n}}+\overline{a_{2}z^{n-1}}+\dots \\
&=\overline{a_{1}z^{n} + a_{2}z^{n-1} +\dots}
\end{align}
$$
$$\begin{align}
\text{We are told that:} \\
a_{1}z^{n} + a_{2}z^{n-1} +\dots+a_{n-1}z^{1}+a_{n}=0 \\
\therefore \overline{a_{1}z^{n} + a_{2}z^{n-1} +\dots+a_{n-1}z^{1}+a_{n}}=\overline{0} \\
\text{But }\overline{0}=0 \text{ (0 has imaginary part of 0)} \\
\therefore a_{1}z^{n} + a_{2}z^{n-1} +\dots=\overline{a_{1}z^{n} + a_{2}z^{n-1} +\dots}=0 \\
\text{(Simplified for space)} \\
\end{align}
$$
$$\begin{align}
\text{We proved above that: } \\
a_{1}\overline{z^{n}} + a_{2}\overline{z^{n-1}} +\dots&=\overline{a_{1}z^{n} + a_{2}z^{n-1} +\dots} \\
\text{And that:} \\
(\overline{z})^{n}&=\overline{z^{n}} \\
\text{Hence,} \\
a_{1}(\overline{z})^{n} + a_{2}(\overline{z})^{n-1} +\dots&= a_{1}\overline{z^{n}} + a_{2}\overline{z^{n-1}} +\dots \\
&=\overline{a_{1}z^{n} + a_{2}z^{n-1} +\dots} \\
&=\overline{0} \\
&=0\text{, as required.} \\
\text{QED}
\end{align}
$$