Question

The complex number z satisfying the equation $\left| {z - i} \right| = \left| {z + 1} \right| = 1$ is (This question has multiple correct options)
$
  (a){\text{ 0}} \\
  (b){\text{ 1 + i}} \\
  (c){\text{ - 1 + i}} \\
  (d){\text{ 1 - i}} \\
 $

Answer 1

Hint: In this question we have to evaluate the equation given, so let the complex number Z satisfying this equation be of the form $x + iy$, where x is the real part and y is the imaginary part. Use the basic definition of $\left| z \right| = x + iy = \sqrt {{x^2} + {y^2}} $ along with the concept mentioned to get the right answer.

Complete step-by-step answer:
It is given z is a complex number so, let
$z = x + iy$
Given equation is
$\left| {z - i} \right| = \left| {z + 1} \right| = 1$
Now substitute the value of z in above equation we have,
$ \Rightarrow \left| {x + iy - i} \right| = \left| {x + iy + 1} \right| = 1$
$ \Rightarrow \left| {x + i\left( {y - 1} \right)} \right| = \left| {\left( {x + 1} \right) + iy} \right| = 1$
Now as we know $\left| {a + ib} \right| = \sqrt {{a^2} + {b^2}} $ so, use this property in above equation we have,
$ \Rightarrow \sqrt {{x^2} + {{\left( {y - 1} \right)}^2}} = \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( y \right)}^2}} = 1$
Now squaring on both sides we have,
$ \Rightarrow {x^2} + {\left( {y - 1} \right)^2} = {\left( {x + 1} \right)^2} + {\left( y \right)^2} = {1^2}$
Now opening the whole square we have,
$ \Rightarrow {x^2} + {y^2} + 1 - 2y = {x^2} + 1 + 2x + {y^2} = 1$…………………. (1)
Now it is also written as
$ \Rightarrow {x^2} + {y^2} + 1 - 2y = {x^2} + 1 + 2x + {y^2}$
Now cancel out common terms we have,
$ \Rightarrow - 2y = 2x$
Divide by 2 we have,
$ \Rightarrow x = - y$ ……………. (2)
From equation (1)
$ \Rightarrow {x^2} + {y^2} + 1 - 2y = 1$
Now substitute the value of x in this equation we have,
$
   \Rightarrow {\left( { - y} \right)^2} + {y^2} + 1 - 2y = 1 \\
   \Rightarrow 2{y^2} - 2y = 0 \\
   \Rightarrow {y^2} - y = 0 \\
   \Rightarrow y\left( {y - 1} \right) = 0 \\
   \Rightarrow y = 0,{\text{ & }}\left( {y - 1} \right) = 0 \\
$
$ \Rightarrow y = 0,1$
Now from equation (2)
$x = - y$
So, if y = 0 $ \Rightarrow x = 0$.
Now, if y = 1 $x = - 1$
So, the complex number
$z = x + iy = 0 + 0i = 0$
And $z = x + iy = - 1 + 1.i = - 1 + i$
Hence, option (a) and (c) is correct.

Note: Whenever we face such type of problems the key point is about the proper simplification of the equation part, since this is a multiple choice problem so be sure that you have evaluated all the possible values coming up for x and y and thus forming different complex numbers which will be satisfying the equation.