# The complex number $z = x + iy$ which satisfy the equation $\left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1$, lie on

${\text{A}}{\text{.}}$ The x-axis

${\text{B}}{\text{.}}$ The straight line $y = 5$

${\text{C}}{\text{.}}$ A circle passing through the origin

${\text{D}}{\text{.}}$ None of these.

Last updated date: 24th Mar 2023

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Answer

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Hint – In this question use the property of modulus of a complex number which is $\left| {A + iB} \right| = \sqrt {{A^2} + {B^2}} $ to reach the answer.

Given equation is

$\left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1$, where $z = x + iy$

Now as we know \[\left| {\dfrac{A}{B}} \right| = \dfrac{{\left| A \right|}}{{\left| B \right|}}\]

$ \Rightarrow \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \dfrac{{\left| {z - 5i} \right|}}{{\left| {z + 5i} \right|}} = 1$

$ \Rightarrow \left| {z - 5i} \right| = \left| {z + 5i} \right|$

Now substitute $z = x + iy$

\[

\Rightarrow \left| {x + iy - 5i} \right| = \left| {x + iy + 5i} \right| \\

\Rightarrow \left| {x + i\left( {y - 5} \right)} \right| = \left| {x + i\left( {y + 5} \right)} \right| \\

\]

Now as we know that $\left| {A + iB} \right| = \sqrt {{A^2} + {B^2}} $, so use this property we have

$\sqrt {{x^2} + {{\left( {y - 5} \right)}^2}} = \sqrt {{x^2} + {{\left( {y + 5} \right)}^2}} $

Now squaring on both sides we have

$

{x^2} + {\left( {y - 5} \right)^2} = {x^2} + {\left( {y + 5} \right)^2} \\

\Rightarrow {\left( {y - 5} \right)^2} = {\left( {y + 5} \right)^2} \\

$

Now opening the square we have

$

{y^2} + 25 - 10y = {y^2} + 25 + 10y \\

\Rightarrow 20y = 0 \\

\Rightarrow y = 0 \\

$

And we all know y = 0 is nothing but a x-axis

Hence option (a) is correct.

Note – In such types of questions the key concept we have to remember is that always recall all the properties of modulus which is stated above, then according to these properties simplify the given equation we will get the required answer.

Given equation is

$\left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = 1$, where $z = x + iy$

Now as we know \[\left| {\dfrac{A}{B}} \right| = \dfrac{{\left| A \right|}}{{\left| B \right|}}\]

$ \Rightarrow \left| {\dfrac{{z - 5i}}{{z + 5i}}} \right| = \dfrac{{\left| {z - 5i} \right|}}{{\left| {z + 5i} \right|}} = 1$

$ \Rightarrow \left| {z - 5i} \right| = \left| {z + 5i} \right|$

Now substitute $z = x + iy$

\[

\Rightarrow \left| {x + iy - 5i} \right| = \left| {x + iy + 5i} \right| \\

\Rightarrow \left| {x + i\left( {y - 5} \right)} \right| = \left| {x + i\left( {y + 5} \right)} \right| \\

\]

Now as we know that $\left| {A + iB} \right| = \sqrt {{A^2} + {B^2}} $, so use this property we have

$\sqrt {{x^2} + {{\left( {y - 5} \right)}^2}} = \sqrt {{x^2} + {{\left( {y + 5} \right)}^2}} $

Now squaring on both sides we have

$

{x^2} + {\left( {y - 5} \right)^2} = {x^2} + {\left( {y + 5} \right)^2} \\

\Rightarrow {\left( {y - 5} \right)^2} = {\left( {y + 5} \right)^2} \\

$

Now opening the square we have

$

{y^2} + 25 - 10y = {y^2} + 25 + 10y \\

\Rightarrow 20y = 0 \\

\Rightarrow y = 0 \\

$

And we all know y = 0 is nothing but a x-axis

Hence option (a) is correct.

Note – In such types of questions the key concept we have to remember is that always recall all the properties of modulus which is stated above, then according to these properties simplify the given equation we will get the required answer.

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