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Question: If \(z = x + iy\) , where x and y are real numbers and \(i = \sqrt { - 1} \) , then the point (x, y)...

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

Explanation

Solution

Hint- In order to solve this question, we have to rationalize z−iz+i\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=−1i = \sqrt { - 1} . These numbers have two parts , one is called as the real part denoted by Re⁡(z)\operatorname{Re} \left( z \right) and the other part is called the imaginary part. Imaginary part is denoted by Im⁡(z)\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.
Given z=x+iy ∴z+i(z−i)=x+iy+ix+iy−i=x+i(y+1)x−i(y−1) =(x+i(y+1))(x−i(1−y))×(x+i(1−y))(x+i(1−y)) =x2+(y2−1)+2ixx+(1−y)2  {\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, z+iz−i\dfrac{{z + i}}{{z - i}} should be purely imaginary
∴Re⁡(z+iz−i)=0 ⇒x2+(y2−1)x2+(1−y)2=0 ⇒x2+y2−1=0 ⇒x2+y2=1  \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 x2+y2=1{x^2} + {y^2} = 1is 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.