Question

The locus of the point z satisfying \[arg\left( {\dfrac{{z - 1}}{{z + 1}}} \right) = k\], (where k is non zero) is [Orissa JEE\[2002\]]
A) Circle with centre on y-axis
B) Circle with centre on x-axis
C) A straight line parallel to x-axis
D) A straight line making an angle \[{60^0}\] with the x-axis

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\[arg\left( {\dfrac{{z - 1}}{{z + 1}}} \right) = k\]
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\[arg\left( {\dfrac{{z - 1}}{{z + 1}}} \right) = k\]
\[arg\left( {\dfrac{{(x + iy) - 1}}{{(x + iy) + 1}}} \right) = k\]
\[arg\left( {\dfrac{{x - 1 + iy}}{{x + 1 + iy}}} \right) = k\]
We know that
\[\arg (\dfrac{a}{b}) = \arg (a) - \arg (b)\]
\[arg(x - 1 + iy) - \arg (x + 1 + iy) = k\]
\[{\tan ^{ - 1}}(\dfrac{y}{{x - 1}}) - {\tan ^{ - 1}}(\dfrac{y}{{x + 1}}) = k\]
\[{\tan ^{ - 1}}[\dfrac{{(\dfrac{y}{{x - 1}}) - \dfrac{y}{{x + 1}}}}{{1 + \dfrac{{{y^2}}}{{({x^2} - 1)}}}}] = k\]
\[\tan (k) = \dfrac{{(y(x + 1) - y(x - 1))}}{{{x^2} + {y^2} - 1}} = \dfrac{{2y}}{{{x^2} + {y^2} - 1}}\]
\[\dfrac{{2y}}{{\tan k}} = {x^2} + {y^2} - 1\]
\[{x^2} + {y^2} - \dfrac{{2y}}{{\tan k}} - 1 = 0\]
\[{x^2} + {y^2} - \dfrac{{2y}}{{\tan k}} - 1 = 0\]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.
Center of circle is given by
\[( - g, - f)\]
Therefore center of above circle is
\[(0,cotk)\]
Clearly center lies on y-axis



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.