Question

Cartesian equation for the locus of z=x+iy in |2z-3|=4

Answer 1

Answer:

$(2x - 3) {}^{2} + y {}^{2} = 16$

Answer 2

Answer:

To form cartesian equation,

we have given |2z-3| =4

But z = x+iy

This implies |2(x+iy) -3| = 4

|2x+2iy-3| = 4

|(2x-3)+ i 2y| = 4

either 2x-3+2y =4 or 2x-3-2y = -4

2x+2y = 7 or 2x-2y = -1