# 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}} \\

$

Last updated date: 25th Mar 2023

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Answer

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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.

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.

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