Question

If \[z = x + iy\;\] and , \[\arg (\dfrac{{z - 2}}{{z + 2}}) = \dfrac{\pi }{6}\] then locus of z is [RPET\[2002\]]
A) A straight line
B) A circle
C) A parabola
D) An ellipse

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\[\arg (\dfrac{{z - 2}}{{z + 2}}) = \dfrac{\pi }{6}\]
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 \[\arg (\dfrac{{z - 2}}{{z + 2}}) = \dfrac{\pi }{6}\]
\[\arg (\dfrac{{(x + iy) - 2}}{{(x + iy) + 2}}) = \dfrac{\pi }{6}\]
We know that
\[\arg \dfrac{b}{a} = \arg (b) - \arg (a)\]
\[\arg ((x + iy) - 2) - \arg ((x + iy) + 2) = \dfrac{\pi }{6}\]
\[{\tan ^{ - 1}}(\dfrac{y}{{x - 2}}) - {\tan ^{ - 1}}(\dfrac{y}{{x + 2}}) = \dfrac{\pi }{6}\]
\[{\tan ^{ - 1}}(\dfrac{{\dfrac{y}{{x - 2}} - \dfrac{y}{{x + 2}}}}{{1 + \dfrac{{{y^2}}}{{{x^2} - 4}}}}) = \dfrac{\pi }{6}\]
\[(\dfrac{{\dfrac{y}{{x - 2}} - \dfrac{y}{{x + 2}}}}{{1 + \dfrac{{{y^2}}}{{{x^2} - 4}}}}) = \tan (\dfrac{\pi }{6})\]
We know that
\[\tan (\dfrac{\pi }{6}) = \dfrac{1}{{\sqrt 3 }}\]
\[(\dfrac{{\dfrac{y}{{x - 2}} - \dfrac{y}{{x + 2}}}}{{1 + \dfrac{{{y^2}}}{{{x^2} - 4}}}}) = \dfrac{1}{{\sqrt 3 }}\]
\[\dfrac{{xy + 2y - xy + 2y}}{{{x^2} + {y^2} - 4}} = \dfrac{1}{{\sqrt 3 }}\]
Now above equation becomes
\[{x^2} + {y^2} - 4\sqrt {3y} - 4 = 0\]
This equation represents the equation of circle.
Here \[{x^2} + {y^2} - 4\sqrt {3y} - 4 = 0\] represent the equation of circle therefore locus of z 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.