Let z[ ]C, the set of complex numbers.

Question

Let z$$ \in $$C, the set of complex numbers. Then the equation, 2|z + 3i| - |z - i| = 0 represents :

Accepted Answer

Given,

2 $$\,\left| \, \right.$$z + 3i$$\,\left| \, \right.$$ = $$\,\left| \, \right.$$z -i$$\,\left| \, \right.$$

Let z = x + iy

$$ \Rightarrow $$ $\,\,\,$2$\sqrt {{x^2} + {{\left( {y + 3} \right)}^2}} $=$\sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} $$ \Rightarrow $$ $\,\,\,$$ 4 (x2 + y2 + 6y + 9) = x2 + y2 - 2y + 1

$$ \Rightarrow $$ $\,\,\,$3x2 + 3y2 + 26y + 35 = 0$ \Rightarrow $$ $\,\,\,$$ x2 + y2 + $${{26} \over 3}$$ y + $${{35} \over 3}$$ = 0

This is a equation of circle with center (- $${{13} \over 3}$$, 0)

$$\therefore\,\,\,$$ Radius = $$\sqrt {0 + {{169} \over 9} - {{35} \over 3}} $$

= $$\sqrt {{{64} \over 9}} $$

= $${8 \over 3}$$

Solutions · 1