Question: If I<sub>m</sub>$\left( \frac{z - 1}{2z + 1} \right)$= –4, then locus of z is-...

Question

If I_m $\left( \frac{z - 1}{2z + 1} \right)$ = –4, then locus of z is-

Answer 1

Sol. Put z = x + iy

Ž I_m $\left( \frac{x - 1 + iy}{2x + 1 + 2iy} \right)$ =–4

= $\frac{- 2y(x - 1) + y(2x + 1)}{(2x + 1)^{2} + 4y^{2}}$ = –4

We get 16x² + 16y² + 16x + 3y + 4 = 0 (Eq. of Circle)

Question: If I<sub>m</sub>\(\left( \frac{z - 1}{2z + 1} \right)\)= –4, then locus of z is-...