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Exercise: Use synthetic division rather than DeMoivre's Theorem to find $(2-i)^7$. Then given $z=2-i$ use synthetic division to find $2z^3-7z^2+5z-3$.

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If one finds the remainder $R(z)$ when $F(z)=z^7$ is divided by

\begin{equation} [z-(2-i)]\cdot[z-(2+i)]=z^2-4z+5 \end{equation}

one finds that

\begin{equation} F(z)=(z^2-4z+5)Q(z)+R(z) \end{equation}

so

\begin{equation} F(2-i)=(2-i)^7=0\cdot Q(2-i)+R(2-i)=-278+29i \end{equation}

For \begin{equation} F(z)=(2z^3-7z^2+5z-3)=(z^2-4z+5)Q(z)+R(z)\\ \end{equation} and \begin{equation} R(z)=-i-8\\ \end{equation} so \begin{equation} F(z)=(z^2-4z+5)Q(z)+R(z)\\ \end{equation} and \begin{equation} F(2-i)=0\cdot Q(2-i)-(2-i)-8=-10+i \end{equation}

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  • $\begingroup$ After $F(z)$ in second line: $-4z$ or $-5z$? $\endgroup$ – imranfat Feb 29 '16 at 3:47
  • $\begingroup$ Yes, you are correct, everywhere I wrote $z^2-5x+5$ it should be $z^2-4z+5$. Thank you for catching that mistake. The remainders are correct, however since I actually used $z^2-4z+5$ in the calculation. $\endgroup$ – John Wayland Bales Mar 4 '16 at 18:22

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