Question

solve it fast please​

Answer 1

Answer:

(4) x + y = 0 is the correct option

Step-by-step explanation:

Given Question :

The locus of the point z = x + y - iy satisfying the equation $\mathrm{\bigg|\dfrac{z-1}{z+1}\bigg| = 1}$ is given by

a) x = 0

b) y = 0

c) x = y

d) x + y = 0

Solution :

Given z = x + y - iy

Condition is

$\longmapsto \mathrm{\bigg|\dfrac{z-1}{z+1}\bigg| = 1}$

$\implies \mathrm{|z -1| = |z + 1|}$

$\implies \mathrm{|x + y - iy -1| = |x+y-iy+1|}$

We know that, If z = a + ib

$\boxed{\mathrm{|z| =|a+ib| = \sqrt{a^2+b^2} }}$

So, rearranging the above terms,

$\implies \mathrm{|(x + y-1) - iy| = |(x+y+1)-iy|}$

Let x + y = t

So,

$\implies \mathrm{|t-1 - iy| = |t+1-iy|}$

$\implies \mathrm{\sqrt{(t-1)^2 + y^2} = \sqrt{(t+1)^2+y^2}}$

Squaring on both sides,

$\implies \mathrm{(t-1)^2 + y^2 = (t+1)^2+y^2}$

$\implies \mathrm{(t-1)^2 + \not{y^2} = (t+1)^2+\not{y^2}}$

$\implies \mathrm{(t-1)^2 = (t+1)^2}$

We know that,

$\boxed{\mathrm{(a + b)^2 = a^2 + 2ab + b^2}}$

$\boxed{\mathrm{(a - b)^2 = a^2 - 2ab + b^2}}$

So, here a = t and b = 1

Thus,

$\implies \mathrm{t^2 -2t + 1= t^2 + 2t - 1}$

$\implies \mathrm{\not{t^2} -2t + \not{1}= \not{t^2} + 2t + \not{1}}$

$\implies \mathrm{-2t = 2t}$

$\implies \mathrm{4t = 0}$

$\implies \mathrm{t = 0}$

We substituted t in place of x + y, so, retrieving them back into the equation, we get,

$\implies \mathrm{x + y = 0}$

$\therefore \small{\underline{\boxed{\bf{\tt{x + y = 0}}}}\ \texttt{is the locus of the point z.} }$

Hope it helps!!