Question

The points of $z$ satisfying $\arg \left( {\dfrac{{z - 1}}{{z + 1}}} \right) = \dfrac{\pi }{4}$ lies on
(a) an arc of a circle
(b) a parabola
(c) an ellipse
(d) a straight line

Answer 1

Hint: We will first let the complex number $z$ be $x + iy$ and then substitute it in the given expression. We will find the argument and therefore, we need the real and complex part of the complex number. Thus, rationalise the expression and apply the formula of the argument of complex numbers and form the equation.

Complete step-by-step answer:
We are given that $\arg \left( {\dfrac{{z - 1}}{{z + 1}}} \right) = \dfrac{\pi }{4}$, where $z$ is a complex number.
Let $z = x + iy$
Then, on substituting the value in the given equation, we will get,
$
  \arg \left( {\dfrac{{x + iy - 1}}{{x + iy + 1}}} \right) = \dfrac{\pi }{4} \\
   \Rightarrow \arg \left( {\dfrac{{x - 1 + iy}}{{x + 1 + iy}}} \right) = \dfrac{\pi }{4} \\
$
We will now rationalise the expression by multiplying and dividing the number by the conjugate of the denominator.
$
  \arg \left( {\left( {\dfrac{{x - 1 + iy}}{{x + 1 + iy}}} \right)\left( {\dfrac{{x + 1 - iy}}{{x + 1 - iy}}} \right)} \right) = \dfrac{\pi }{4} \\
   \Rightarrow \arg \left( {\dfrac{{{x^2} + x - ixy - x - 1 + iy + ixy + iy + {y^2}}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}} \right) = \dfrac{\pi }{4} \\
   \Rightarrow \arg \left( {\dfrac{{{x^2} - 1 + {y^2} + i2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}} \right) = \dfrac{\pi }{4} \\
$
Which can also be written as \[\arg \left( {\dfrac{{{x^2} - 1 + {y^2}}}{{{{\left( {x + 1} \right)}^2} + {y^2}}} + i\dfrac{{2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}} \right) = \dfrac{\pi }{4}\]
Also, we know that if the complex number is of the form $a + ib$, then argument of the complex number is given by ${\tan ^{ - 1}}\left( {\dfrac{b}{a}} \right)$
Therefore, from the above equation, we have,
$
  {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{2y}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}}}{{\dfrac{{{x^2} - 1 + {y^2}}}{{{{\left( {x + 1} \right)}^2} + {y^2}}}}}} \right) = \dfrac{\pi }{4} \\
   \Rightarrow \dfrac{{2y}}{{{x^2} - 1 + {y^2}}} = \tan \dfrac{\pi }{4} \\
   \Rightarrow \dfrac{{2y}}{{{x^2} - 1 + {y^2}}} = 1 \\
$
On cross-multiplying, we will have,
$
  2y = {x^2} - 1 + {y^2} \\
   \Rightarrow {x^2} - 1 + {y^2} - 2y = 0 \\
$
The above equation represents an equation of a circle.
Therefore, the points of $z$ satisfying the condition $\arg \left( {\dfrac{{z - 1}}{{z + 1}}} \right) = \dfrac{\pi }{4}$ lies on the arc of a circle.

Hence, option (a) is correct.

Note: The argument of the complex number is the angle formed by the complex number with the real axis or $x$ axis . It is measured in radians. Also, the modulus of the complex number gives the magnitude of the complex number.