We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
83
1
avatar

thanks

 Apr 15, 2019
 #1
avatar+22883 
+1

​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 \(\mathbf{z^2+8z+20 = 0}\) .

 

laugh

 Apr 16, 2019

6 Online Users