# 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

Verified

381.6k+ views

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.

Recently Updated Pages

Define absolute refractive index of a medium

Find out what do the algal bloom and redtides sign class 10 biology CBSE

Prove that the function fleft x right xn is continuous class 12 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Find the values of other five trigonometric ratios class 10 maths CBSE

Find the values of other five trigonometric functions class 10 maths CBSE

Trending doubts

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference Between Plant Cell and Animal Cell

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Give 10 examples for herbs , shrubs , climbers , creepers

Which planet is known as the red planet aMercury bMars class 6 social science CBSE

Which state has the longest coastline in India A Tamil class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE