Question

The locus of the points \[z\] which satisfy the condition arg \[(\dfrac{{z - 1}}{{z + 1}}) = \dfrac{\pi }{3}\]is
A) A straight line
B) A circle
C) A Parabola
D) None of these

Answer 1

Hint: in this question we have to find locus of the point \[z\] which satisfy the given condition. First write the given complex number as a combination of real and imaginary number. Then apply formula for argument.



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
\[\arg (z) = {\tan ^{ - 1}}(\dfrac{y}{x})\]



Complete step by step solution:Given: Argument of complex number is given
Now we have argument which is equal to arg \[(\dfrac{{z - 1}}{{z + 1}}) = \dfrac{\pi }{3}\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number

Put this value in \[(\dfrac{{z - 1}}{{z + 1}})\]
\[\dfrac{{(x + iy) - 1}}{{(x + iy) + 1}}\]
\[\dfrac{{({x^2} + {y^2} - 1) + 2iy}}{{{{(x + 1)}^2} + {y^2}}}\]
\[arg(\dfrac{{z - 1}}{{z + 1}}) = {\tan ^{ - 1}}(\dfrac{{2y}}{{{x^2} + {y^2} - 1}})\]
It is given in the question that arg\[(\dfrac{{z - 1}}{{z + 1}}) = \dfrac{\pi }{3}\]
\[{\tan ^{ - 1}}(\dfrac{{2y}}{{{x^2} + {y^2} - 1}}) = \dfrac{\pi }{3}\]
\[(\dfrac{{2y}}{{{x^2} + {y^2} - 1}}) = \tan (\dfrac{\pi }{3})\]
We know that \[\tan (\dfrac{\pi }{3}) = \sqrt 3 \]
\[(\dfrac{{2y}}{{{x^2} + {y^2} - 1}}) = \sqrt 3 \]
\[{x^2} + {y^2} - 1 = \dfrac{{2y}}{{\sqrt 3 }}\]
\[{x^2} + {y^2} - \dfrac{{2y}}{{\sqrt 3 }} - 1 = 0\] 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 \[{x^2} + {y^2} - \dfrac{{2y}}{{\sqrt 3 }} - 1 = 0\] 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.