Question

If $z = x + iy$ , where x and y are real numbers and $i = \sqrt { - 1} $ , then the point (x, y)
For which $\dfrac{{z - i}}{{z + i}}$ is real, lie on
(A). An ellipse
(B). A circle
(C).A parabola
(D).A straight line

Answer 1

Hint- In order to solve this question, we have to rationalize $\dfrac{{z - i}}{{z + i}}$ since we know that it is purely imaginary. Thus, equating the real part = 0. We get our desired equation.

Complete step-by-step solution -
Well first one should know that this is a concept of algebra of complex numbers. The first basic concept is what is the algebra of complex numbers. First know that complex numbers are an algebraic expression including the factor $i = \sqrt { - 1} $. These numbers have two parts , one is called as the real part denoted by $\operatorname{Re} \left( z \right)$ and the other part is called the imaginary part. Imaginary part is denoted by $\operatorname{Im} \left( z \right)$ for the complex number represented by ‘z’. Also one should know another important thing: the equation of circle which is basically the main part of the question.
 $
  {\text{Given }}z = x + iy \\
  \therefore \dfrac{{z + i}}{{\left( {z - i} \right)}} = \dfrac{{x + iy + i}}{{x + iy - i}} = \dfrac{{x + i\left( {y + 1} \right)}}{{x - i\left( {y - 1} \right)}} \\
   = \dfrac{{\left( {x + i\left( {y + 1} \right)} \right)}}{{\left( {x - i\left( {1 - y} \right)} \right)}} \times \dfrac{{\left( {x + i\left( {1 - y} \right)} \right)}}{{\left( {x + i\left( {1 - y} \right)} \right)}} \\
   = \dfrac{{{x^2} + \left( {{y^2} - 1} \right) + 2ix}}{{x + {{\left( {1 - y} \right)}^2}}} \\
 $ …..(i) (by cross multiplying)
Since, $\dfrac{{z + i}}{{z - i}}$ should be purely imaginary
$
  \therefore \operatorname{Re} \left( {\dfrac{{z + i}}{{z - i}}} \right) = 0 \\
   \Rightarrow \dfrac{{{x^2} + \left( {{y^2} - 1} \right)}}{{{x^2} + {{\left( {1 - y} \right)}^2}}} = 0 \\
   \Rightarrow {x^2} + {y^2} - 1 = 0 \\
   \Rightarrow {x^2} + {y^2} = 1 \\
 $
Now, as we can see that ${x^2} + {y^2} = 1$is an equation of circle
Hence, (x, y) lie on circle
Therefore, (B) is the correct option.

Note- Here in this question, one can make a mistake in the cross-multiplying equation (i) and in the simplification part that is, after putting real part = 0. By these basics we will be able to solve this question.