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​complex roots

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thanks

Apr 15, 2019

#1
+25644
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​complex roots

$$\begin{array}{|rcll|} \hline w &=& p+2i \\ w^{*} &=& p-2i \\\\ \left[ z-w \right] \left[z-w^{*} \right] &=& z^2+(4+i+qi)z + 20 \\ \left[ z-(p+2i) \right] \left[z-(p-2i) \right] &=& z^2+(4+i+qi)z + 20 \\ \left[ (z-p)-2i) \right] \left[(z-p)+2i) \right] &=& z^2+(4+i+qi)z + 20 \\ (z-p)^2-(2i)^2 &=& z^2+(4+i+qi)z + 20 \\ (z-p)^2+4 &=& z^2+(4+i+qi)z + 20 \\ z^2-2pz+(p^2+4) &=& z^2+\underbrace{(4+i+qi)}_{=-2p}z + \underbrace{20}_{=p^2+4} \\\\ p^2+4 &=& 20 \\ p^2 &=& 16 \\ \mathbf{p} & \mathbf{=} & \mathbf{\pm 4} \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline p=4: & 4+i+qi &=& -2p \\ & 4+i+qi &=& -8 \\ & qi &=& -12-i \\ & q &=& \dfrac{-12-i}{i} \\ & q &=& \dfrac{(-12-i)}{i}\cdot \dfrac{i}{i} \\ & q &=& \dfrac{(-12-i)i}{i^2} \quad | \quad i^2 = -1 \\ & q &=& \dfrac{(-12-i)i}{-1} \\ & q &=& (12+i)i \\ & q &=& i^2+12i \\ & \mathbf{q} & \mathbf{=} & \mathbf{-1+12i} \\\\ & w &=& 4+2i \\ & w^* &=& 4-2i \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline p=-4: & 4+i+qi &=& -2p \\ & 4+i+qi &=& 8 \\ & qi &=& 4-i \\ & q &=& \dfrac{4-i}{i} \\ & q &=& \dfrac{(4-i)}{i}\cdot \dfrac{i}{i} \\ & q &=& \dfrac{(4-i)i}{i^2} \quad | \quad i^2 = -1 \\ & q &=& \dfrac{(4-i)i}{-1} \\ & q &=& (i-4)i \\ & q &=& i^2-4i \\ & \mathbf{q} & \mathbf{=} & \mathbf{-1-4i} \\\\ & w &=& -4+2i \\ & w^* &=& -4-2i \\ \hline \end{array}$$

The possible values of $$\mathbf{q}$$ are $$\mathbf{-1+12i}$$ and $$\mathbf{-1-4i}$$

and the quadratic equations are $$\mathbf{z^2-8z+20 = 0}$$ and $$\mathbf{z^2+8z+20 = 0}$$ .

Apr 16, 2019