
When \[\dfrac{{z + i}}{{z + 2\;}}\] is purely imaginary, the locus described by the point \[z\] in the Argand diagram is
A) Circle of radius \[\dfrac{{\sqrt 5 }}{2}\]
B) Circle of radius \[\dfrac{5}{4}\]
C) Straight line
D) Parabola
Answer
219.3k+ views
Hint: in this question we have to find locus described by the point \[z\] in the Argand diagram representing what shape. First write the given complex number as a combination of real and imaginary number. Put z in form of real and imaginary number into the equation.
Formula Used:\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one
Complete step by step solution:Given: A complex number and it is given that complex number is purely imaginary
Now we have complex number\[\dfrac{{z + i}}{{z + 2\;}}\]
We know that
\[z = x + iy\]
Put this value in\[\dfrac{{z + i}}{{z + 2\;}}\]
\[\dfrac{{(x + iy) + i}}{{(x + iy) + 2\;}}\]
\[\dfrac{{x + iy + i}}{{x + iy + 2\;}} = \dfrac{{x + i(y + 1)}}{{(x + 2) + iy\;}}\]
\[\dfrac{{x + i(y + 1)}}{{(x + 2) + iy\;}} = \dfrac{{[x + i(y + 1)][(x + 2) - iy]}}{{[x + 2) + iy][(x + 2) - iy]}}\]
\[[\dfrac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}}] + i\dfrac{{(y + 1)(x + 2) - xy}}{{{{(x + 2)}^2} + {y^2}}}\]
It is given in the question that complex number is purely imaginary, so its real part will be zero
\[\dfrac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}} = 0\]
\[{x^2} + 2x + {y^2} + y = 0\]
This equation represents the circle. \[{{\bf{x}}^{\bf{2}}}\; + {\rm{ }}{{\bf{y}}^{\bf{2}}}\; + {\rm{ }}{\bf{2gx}}{\rm{ }} + {\rm{ }}{\bf{2fy}}{\rm{ }} + {\rm{ }}{\bf{c}}{\rm{ }} = {\rm{ }}{\bf{0}}\]
Radius of circle is given by
\[\sqrt {{g^2} + {f^2} - c} \]
Radius of circle is \[\sqrt {1 + \dfrac{1}{4} - 0} = \dfrac{{\sqrt 5 }}{2}\]
Here \[{x^2} + 2x + {y^2} + y = 0\]represent the equation of circle therefore locus of point represent circle.
Option ‘A’ is correct
Note: Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.
Formula Used:\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one
Complete step by step solution:Given: A complex number and it is given that complex number is purely imaginary
Now we have complex number\[\dfrac{{z + i}}{{z + 2\;}}\]
We know that
\[z = x + iy\]
Put this value in\[\dfrac{{z + i}}{{z + 2\;}}\]
\[\dfrac{{(x + iy) + i}}{{(x + iy) + 2\;}}\]
\[\dfrac{{x + iy + i}}{{x + iy + 2\;}} = \dfrac{{x + i(y + 1)}}{{(x + 2) + iy\;}}\]
\[\dfrac{{x + i(y + 1)}}{{(x + 2) + iy\;}} = \dfrac{{[x + i(y + 1)][(x + 2) - iy]}}{{[x + 2) + iy][(x + 2) - iy]}}\]
\[[\dfrac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}}] + i\dfrac{{(y + 1)(x + 2) - xy}}{{{{(x + 2)}^2} + {y^2}}}\]
It is given in the question that complex number is purely imaginary, so its real part will be zero
\[\dfrac{{{x^2} + 2x + {y^2} + y}}{{{{(x + 2)}^2} + {y^2}}} = 0\]
\[{x^2} + 2x + {y^2} + y = 0\]
This equation represents the circle. \[{{\bf{x}}^{\bf{2}}}\; + {\rm{ }}{{\bf{y}}^{\bf{2}}}\; + {\rm{ }}{\bf{2gx}}{\rm{ }} + {\rm{ }}{\bf{2fy}}{\rm{ }} + {\rm{ }}{\bf{c}}{\rm{ }} = {\rm{ }}{\bf{0}}\]
Radius of circle is given by
\[\sqrt {{g^2} + {f^2} - c} \]
Radius of circle is \[\sqrt {1 + \dfrac{1}{4} - 0} = \dfrac{{\sqrt 5 }}{2}\]
Here \[{x^2} + 2x + {y^2} + y = 0\]represent the equation of circle therefore locus of point represent circle.
Option ‘A’ is correct
Note: Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.
Recently Updated Pages
JEE Main 2022 (July 27th Shift 2) Maths Question Paper with Answer Key

JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2023 (April 6th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 June 29 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

Hybridization of Atomic Orbitals Important Concepts and Tips for JEE

Trending doubts
Understanding Collisions: Types and Examples for Students

Understanding Atomic Structure for Beginners

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Electromagnetic Waves and Their Importance

Other Pages
Understanding Average and RMS Value in Electrical Circuits

Understanding Entropy Changes in Different Processes

Common Ion Effect: Concept, Applications, and Problem-Solving

Understanding Excess Pressure Inside a Liquid Drop

Understanding Elastic Collisions in Two Dimensions

Devuthani Ekadashi 2025: Correct Date, Shubh Muhurat, Parana Time & Puja Vidhi

