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