Question

: If \[\left| {z + 1} \right| = \sqrt 2 \left| {z - 1} \right|\]then the locus described by the point \[z\] in the Argand diagram is a
A) Straight line
B) Circle
C) Parabola
D) None of these

Answer 1

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:Equation of complex number is given by
\[z = x + iy\]
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 equation
Now we have complex equation equal to \[\left| {z + 1} \right| = \sqrt 2 \left| {z - 1} \right|\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Put this value in complex equation\[\left| {z + 1} \right| = \sqrt 2 \left| {z - 1} \right|\]
Now we have
\[\left| {x + iy + 1} \right| = \sqrt 2 \left| {x + iy - 1} \right|\]
\[\left| {(x + 1) + iy} \right| = \sqrt 2 \left| {(x - 1) + iy} \right|\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
After putting value of modulus in equation\[\left| {(x + 1) + iy} \right| = \sqrt 2 \left| {(x - 1) + iy} \right|\]we get
\[{(x + 1)^2} + {y^2} = 2[{(x - 1)^2} + {y^2}]\]
\[{x^2} + {y^2} - 6x + 1 = 0\]it is in the form of \[{{\bf{x}}^{\bf{2}}}\; + {\rm{ }}{{\bf{y}}^{\bf{2}}}\; + {\rm{ }}{\bf{2gx}}{\rm{ }} + {\rm{ }}{\bf{2fy}}{\rm{ }} + {\rm{ }}{\bf{c}}{\rm{ }} = {\rm{ }}{\bf{0}}\]
This equation represents the circle.
Here \[{x^2} + {y^2} - 6x + 1 = 0\] represent the equation of circle therefore locus of point represent circle.



Option ‘B’ 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.