If |z| = |z-i/3| then z lies on what and how ?

2 Answers
Mar 26, 2018

The answer is #y=1/6#

Explanation:

Let #z=x+iy#

Then,

#|z|=|z-1/3i|#

#|x+iy|=|x+iy-1/3i|#

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

Therefore,

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

#cancelx^2+y^2=cancelx^2+(y-1/3)^2#

#cancely^2=cancely^2-2/3y+1/9#

#2/3y=1/9#

#y=1/6#

The locus of #z# lies on a line parallel to the #"x-axis"# and the equation is #y=1/6#

Mar 26, 2018

#z# lies on a circle having centre at #(0,-1/8)=-1/8i# and radius #3/8#.

Explanation:

Hoping that the Problem is to find the Locus of #z# if, #|z|=|(z-i)/3|#.

Let, #z=x+iy; x,y in RR. :. z-i=x+i(y-1)#.

# rArr |z|=sqrt(x^2+y^2), and, |z-i|=sqrt{x^2+(y-1)^2}#.

# :." Given that "|z|=|(z-i)/3|=|(z-i)|/3#,

#rArr sqrt(x^2+y^2)=1/3sqrt{x^2+(y-1)^2}#.

# rArr 9(x+y^2)=9x^2+9y^2=x^2+y^2-2y+1#.

# rArr 8x^2+8y^2+2y=1, or, x^2+y^2+y/4=1/8#.

#:. x^2+y^2+2*y*1/8+1/8^2=1/8^2+1/8=(1+8)/8^2#.

#:. x^2+(y+1/8)^2=(3/8)^2, i.e., #

# (x-0)^2+(y-(-1/8))^2=(3/8)^2#, which shows that

#(x,y), or, z=x+iy# lies on a circle having centre at

#(0,-1/8)=-1/8i# and radius #3/8#.