Question

|z+i| = |z-i|
Then find the locus of 'z'.

Answer 1

if you put z=a+bi then
\[ \large |z+i|=|z-i| \]
\[ \large |a+(b+1)i|=|a+(b-1)i| \]
\[ \large \sqrt{a^2+(b+1)^2}=\sqrt{a^2+(b-1)^2} \]
\[ \large a^2+b^2+2b+1=a^2+b^2-2b+1 \]
can you go on?

Answer 2

ya so y axis is zero
then
it means that locus is x-axis ??????????????????

Answer 3

square both sides can prevent the negetive situation

Answer 4

no the other way around