If # z in C#, then what does the equation #2|z+3i|-|z-i|=0# represent?

Question

4749 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T16:17:37

This is the equation of a circle, center #(0, -13/3)# and radius #=8/3#

The equation is

#2|z+3i|=|z-i|#

The modulus of #(z-i)# is twice the modulus of #(z+3i)#

Let #z=x+iy#

Then,

#2|x+iy+3i|=|x+iy-i|#

#2|x+i(y+3)|=|x+i(y-1)|#

Then,

#2sqrt(x^2+(y+3)^2)=sqrt(x^2+(y-1)^2)#

Squaring both sides

#4(x^2+(y+3)^2)=(x^2+(y-1)^2)#

#4(x^2+y^2+6y+9)=(x^2+y^2-2y+1)#

#4x^2-x^2+4y^2-y^2+24y+2y+36-1#

#3x^2+3y^2+26y+35=0#

#=3x^2+3(y^2+26/3y+169/9)=-35+3*169/9#

#=x^2+(y+13/3)^2=64/9=(8/3)^2#

This is the equation of a circle, center #(0, -13/3)# and radius #=8/3#

graph{x^2+(y+13/3)^2-64/9=0 [-7.6, 10.18, -7.55, 1.34]}