​Q.3

​Q.3 integer function). _sm

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

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \left|\frac{3\mathrm{z}-\mathrm{i}}{\mathrm{z}+2}\right|=\mathrm{\lambda } \\ \Rightarrow \frac{\left|3\mathrm{z}-\mathrm{i}\right|}{\left|\mathrm{z}+2\right|}=\mathrm{\lambda } \\ \Rightarrow \left|3\mathrm{z}-\mathrm{i}\right|=\mathrm{\lambda }\left|\mathrm{z}+2\right| \\ \Rightarrow 3\left|\mathrm{z}-\frac{\mathrm{i}}{3}\right|=\mathrm{\lambda }\left|\mathrm{z}+2\right| \\ \Rightarrow \left|\mathrm{z}-\frac{\mathrm{i}}{3}\right|=\frac{\mathrm{\lambda }}{3}\left|\mathrm{z}+2\right| \\ \mathrm{Now} \mathrm{loucs} \mathrm{depends} \mathrm{on} \mathrm{\lambda }, \mathrm{if} \mathrm{\lambda }=3 \\ \Rightarrow \left|\mathrm{z}-\frac{\mathrm{i}}{3}\right|=\left|\mathrm{z}+2\right| \\ \mathrm{Which} \mathrm{is} \mathrm{equation} \mathrm{of} \mathrm{perpendicular} \mathrm{bisector} \mathrm{of} \mathrm{line} \mathrm{joining} \frac{\mathrm{i}}{3} \mathrm{and} -2 \\ \mathrm{Hence} \mathrm{locus} \mathrm{can} \mathrm{be}\mathbf{ }\mathbf{straight}\mathbf{ }\mathbf{line}\mathbf{.} \\ \mathrm{Similarly} \mathrm{for} \mathrm{\lambda }=6 \mathrm{locus} \mathrm{will} \mathrm{be} \mathbf{circle}\mathbf{.} \end{gathered}$$



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