Do not use a calculator in answering this question. Explain why the equation ${{z}^{3}}+a{{z}^{2}}+a

Question

Do not use a calculator in answering this question. Explain why the equation ${{z}^{3}}+a{{z}^{2}}+az+7=0$ cannot have more than two non-real roots, where $a$ is a real constant. [1] Given that $z=-7$ is a root of the equation in (i), find the other roots, leaving your answers in the form $r{{e}^{\mathrm{i}	heta }}$, where $r>0$ and $-\pi [4] Hence, solve the equation $\mathrm{i}{{z}^{3}}+8{{z}^{2}}-8\mathrm{i}z-7=0$, leaving your answers in the form $r{{e}^{\mathrm{i}	heta }}$, where $r>0$ and $-\pi [2]

Accepted Answer

(ii) $a=8$; other roots $e^{2\pi i/3}$, $e^{-2\pi i/3}$; (iii) $7e^{i\pi/2},e^{i\pi/6},e^{i5\pi/6}$

2017 NYJC P1 Q3

2017 NYJC P1 Q3

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