Verify that one of the roots of the equation \({{z}^{3}}-\left( 1+2\mathrm{i} \right){{z}^{2}}+\left

Question

Verify that one of the roots of the equation ${{z}^{3}}-\left( 1+2\mathrm{i} \right){{z}^{2}}+\left( a-1+\mathrm{i} \right)z-a\left( 1+\mathrm{i} \right)=0$ where $a$ is real, is $1+\mathrm{i}$.[2] Show that the other $2$ roots ${{z}_{1}}$ and ${{z}_{2}}$ can be expressed as ${{z}_{1}}=\frac{\sqrt{-1-4a}+\mathrm{i}}{2}$ and ${{z}_{2}}=\frac{-\sqrt{-1-4a}+\mathrm{i}}{2}$.[3] Find the range of $a$ such that ${{z}_{1}}$ and ${{z}_{2}}$ are purely imaginary.[2] Given that $\arg \left( {{z}_{1}} \right)=\frac{\pi }{3}$, find $a$.[3] Hence find $\left| {{z}_{2}} \right|$.[2]

Accepted Answer

(i) $1+i$ is a root $(\text{verified})$ (ii) $z_1=\frac{\sqrt{-1-4a}+i}{2}$, $z_2=\frac{-\sqrt{-1-4a}+i}{2}$ (iii) $a\ge-\frac14$ (iv) $a=-\frac13$, $|z_2|=\frac{\sqrt3}{3}$

2021 TJC P1 Q9

2021 TJC P1 Q9

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