Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

Get your **free** account and access **expert** answers to this and **thousands** of other questions

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

i think it is a l ine

- anonymous

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,

- anonymous

@satellite73 rethink please :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

- anonymous

scratch that, i think it is a region bounded by two line

- anonymous

OK so since it is IIT based so I will say : There may be no answers or multiple answers

- anonymous

Well yes @satellite73 has the answer, it will be a line \(\textbf{segment}\)

- anonymous

But I am half way stuck @satellite73 , can you show your work, I think @experimentX is also writing his work :)

- experimentX

put z = x + iy
you would get,
\[ \sqrt{(y-1)^2 + x^2} + \sqrt{x^2 + (y-1)^2} = 2\]
http://www.wolframalpha.com/input/?i=plot+\sqrt{%28y-1%29^2+%2B+x^2}+%2B+\sqrt{x^2+%2B+%28y-1%29^2}+%3D+2
guess it is a circle

- experimentX

woops!! (y+1)^2

- experimentX

http://www.wolframalpha.com/input/?i=plot+%7C+sqrt%28%28y%2B1%29%5E2%2Bx%5E2%29%2Bsqrt%28x%5E2%2B%28y-1%29%5E2%29+%3D+2

- anonymous

What I can do is :
\[\large{|(-iz-1)| + |(z-i)| =2}\]
\[\large{|i(z+i)|+|z-i|=2}\]
\[\large{|i||z+i| + |z-i| = 2 }\]
\[\large{|z+i| + |z-i| = 2}\]

- anonymous

that looks good

- anonymous

|dw:1352908224557:dw|

- anonymous

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! )

- anonymous

yes \(x=0\)

- anonymous

i did it the donkey way but your way is much much better i think

- UnkleRhaukus

\(z=i\) is a solution

- anonymous

Oh! No problem, sometimes we learn a lot from "that say donkey ways" , thanks for your time : @experimentX @satellite73

- experimentX

|z| <= 1 and z = i

- experimentX

no wonder it does not show in x's and y's

- UnkleRhaukus

\(z=-i\) is also a solution

- experimentX

the solution is a line segment from +i to -i

- UnkleRhaukus

\[\large{|z+i| + |z-i| = 2}\]
change of variables
shift along the complex plane
\[w=z+i\]
\[{|w| + |w-2i| = 2}\]

- UnkleRhaukus

|dw:1352909335544:dw|

- UnkleRhaukus

circle + circle = circle?

Looking for something else?

Not the answer you are looking for? Search for more explanations.