Question

Absolute Value question

Answer 1

Assuming \(|x+iy-1| = |x+iy-2i|\), express \(y\) in terms of \(x\)
Where \(x,y\in\mathbb{R}\)

Answer 2

Any idea, @ganeshie8?

Answer 3

z is a complex number equidistant from (1,0) and (0,2)
so it is a straight line perpendicular to the line joining both of them and passing through the mid point
you will get the relation between y and x

Answer 4

Nice!

Answer 5

I am confused. What do you mean by \(\mathbf{equidistant }\)?

Answer 6

let \(z=x+iy\)

then the given equation is same as
\[|z-1| = |z-2i|\]

Answer 7

recall that |z-1| gives you the distance between \(z\) and \(1+0i\)

and

|z-2i| gives the distance between \(z\) and \(0+2i\)

Answer 8

if you don't like that geometric thingy, you may work it algebraically :

\[|x+iy-1| = |x+iy-2i|\\~\\|(x-1)+iy|=|x+i(y-2)|\\~\\(x+1)^2+y^2 = x^2+(y-2)^2\]

simplify

Answer 9

May I ask, why you squared everything?

Answer 10

NVM, REALISED

Answer 11

I get \[y=\dfrac{2x}{4}+\dfrac{3}{4}\]. Is that right?

Answer 12

Wait, no

Answer 13

Why did:
\[|(x−1)+iy|=|x+i(y−2)| \]Become\[
(x+1)^2+y^2=x^2+(y−2)^2\], Instead of\[
(x-1)^2+y^2=x^2+(y−2)^2\]

Answer 14

Ahh thats just my mistake!

Answer 15

Ahh, phew ;)

Answer 16

The answer I get is \[\huge{y=\dfrac{x}{2}+\dfrac{3}{4}}\]

Answer 17

The answer in the book is\[\huge{y=\dfrac{x}{4}+\dfrac{3}{4}}\]

Answer 18

What do you think is the issue, @ganeshie8?

Answer 19

y = x/2 + 3/4 is right

Answer 20

your textbook must be having a typo

Answer 21

Ahh, ok then. Thanks so much!. Could you help me with another question?

Answer 22

wil try, pls post as a new q so that it gets bumped in the top and others see

Answer 23

Ok then :D