Question

z-2 43. If z= x + iy and = 2, show that 3(x2 + y2) – 20x + 32 = 0. ​

Answer 1

Answer:

$| \frac{z - 2}{z - 3} | = 2$

$|z - 2| = 2 \: |z - 3|$

$|z - {2}|^{2} = 4 |z - {3}|^{2}$

Put z=x+iy

→∣(x−2)+iy∣2=4∣(x−3)+iy∣2

→(x−2)2+y2=4((x−3)2+y2)

→x2−4x+4+y2=4(x2−6x+9+y2)

→x2+y2−4x+4=4(x2+y2)−24x+36

→3(x2+y2)−20x+32=0