let z1 and z2 be two non-real complex cube roots of unity lz-z1l^2+lz-z2l^2= lamda be the equation of a circle with z1, z2 as the ends of diameter then the value of lamda is
 

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} {\mathrm{z}}_1 \mathrm{and} {\mathrm{z}}_2 \mathrm{are} \mathrm{roots} \mathrm{of} \mathrm{unity}, \mathrm{hence} \\ {\mathrm{z}}_1=-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i} \\ {\mathrm{z}}_2=-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i} \\ \mathrm{As} {\mathrm{z}}_1,{\mathrm{z}}_2 \mathrm{are} \mathrm{ends} \mathrm{of} \mathrm{diameter}, \mathrm{hence} \mathrm{equation} \mathrm{of} \mathrm{circle} \mathrm{is} \\ \left(\mathrm{x}+\frac{1}{2}\right)\left(\mathrm{x}+\frac{1}{2}\right)+\left(\mathrm{y}-\frac{\sqrt{3}}{2}\right)\left(\mathrm{y}+\frac{\sqrt{3}}{2}\right)=0 \\ \Rightarrow {\mathrm{x}}^2+\frac{1}{4}+\mathrm{x}+{\mathrm{y}}^2-\frac{3}{4}=0 \\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}+{\mathrm{y}}^2-\frac{1}{2}=0 \\ \Rightarrow 2{\mathrm{x}}^2+2\mathrm{x}+2{\mathrm{y}}^2-1=0 \\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}+{\mathrm{y}}^2+{\mathrm{x}}^2+\mathrm{x}+{\mathrm{y}}^2-1=0 \\ \Rightarrow {\left(\mathrm{x}+\frac{1}{2}\right)}^2+{\left(\mathrm{y}-\frac{\sqrt{3}}{2}\right)}^2+{\left(\mathrm{x}+\frac{1}{2}\right)}^2+{\left(\mathrm{y}+\frac{\sqrt{3}}{2}\right)}^2-\frac{1}{4}-\frac{1}{4}-\frac{3}{4}-\frac{3}{4}-1=0 \\ \Rightarrow {\left(\mathrm{x}+\frac{1}{2}\right)}^2+{\left(\mathrm{y}-\frac{\sqrt{3}}{2}\right)}^2+{\left(\mathrm{x}+\frac{1}{2}\right)}^2+{\left(\mathrm{y}+\frac{\sqrt{3}}{2}\right)}^2-3=0 \\ \Rightarrow {\left(\mathrm{x}+\frac{1}{2}\right)}^2+{\left(\mathrm{y}-\frac{\sqrt{3}}{2}\right)}^2+{\left(\mathrm{x}+\frac{1}{2}\right)}^2+{\left(\mathrm{y}+\frac{\sqrt{3}}{2}\right)}^2=3 \\ \Rightarrow {\left|\mathrm{z}-{\mathrm{z}}_1\right|}^2+{\left|\mathrm{z}-{\mathrm{z}}_2\right|}^2=3 \\ \Rightarrow \mathrm{\lambda }=3 \left[\mathrm{Answer}\right] \end{gathered}$$

Hope this information will clear your doubts about this topic.

If you have any doubts just ask here on the ask and answer forum and our experts will try to help you out as soon as possible.
Regards