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Let z be a complex number than locus represented by |iz-1| + |z-i| = 2 is : a) a line , b) a circle , c) a pair of straight lines, d) a parabola

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i think it is a l ine
Sorry for the mistake in the question earlier , the modified quest. is in the post itself. I seek help from the users presented here as soon as possible,
@satellite73 rethink please :)

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Other answers:

scratch that, i think it is a region bounded by two line
OK so since it is IIT based so I will say : There may be no answers or multiple answers
Well yes @satellite73 has the answer, it will be a line \(\textbf{segment}\)
But I am half way stuck @satellite73 , can you show your work, I think @experimentX is also writing his work :)
put z = x + iy you would get, \[ \sqrt{(y-1)^2 + x^2} + \sqrt{x^2 + (y-1)^2} = 2\]\sqrt{%28y-1%29^2+%2B+x^2}+%2B+\sqrt{x^2+%2B+%28y-1%29^2}+%3D+2 guess it is a circle
woops!! (y+1)^2
What I can do is : \[\large{|(-iz-1)| + |(z-i)| =2}\] \[\large{|i(z+i)|+|z-i|=2}\] \[\large{|i||z+i| + |z-i| = 2 }\] \[\large{|z+i| + |z-i| = 2}\]
that looks good
So am I right in my way @experimentX @satellite73 ? (Since I am only given 54 seconds to do these type of questions so I prefer to use shortcuts that is here : not putting z = x + iy , my opinion said , yes ! x will be equal to zero if we are going to solve it! )
yes \(x=0\)
i did it the donkey way but your way is much much better i think
\(z=i\) is a solution
Oh! No problem, sometimes we learn a lot from "that say donkey ways" , thanks for your time : @experimentX @satellite73
|z| <= 1 and z = i
no wonder it does not show in x's and y's
\(z=-i\) is also a solution
the solution is a line segment from +i to -i
\[\large{|z+i| + |z-i| = 2}\] change of variables shift along the complex plane \[w=z+i\] \[{|w| + |w-2i| = 2}\]
circle + circle = circle?

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