Question

If a complex number $\dfrac{{z + 1}}{{z + i}}$ is purely imaginary, then z lies on a
A) Straight line
B) Circle
C) Circle with radius 1
D) Circle passing through (1,1)

Answer 1

Hint: First of all take z as an complex number by letting it in the form of $x + iy$ As it is given that the number is purely imaginary then take the real part as 0, and solve it till the end you will get an equation in x and y observe and see which one of the above options satisfies.
Complete Step by Step Solution:
We are given that $\dfrac{{z + 1}}{{z + i}}$ is purely imaginary.
${\mathop{\rm Re}\nolimits} \left( {\dfrac{{z + 1}}{{z + i}}} \right) = 0$
Let us assume that $z = x + iy$ then the above thing can be written as
${\mathop{\rm Re}\nolimits} \left( {\dfrac{{x + iy + 1}}{{x + iy + i}}} \right) = 0$
So from here we can write it as

\therefore {\mathop{\rm Re}\nolimits} \left( {\dfrac{{\left( {x + 1} \right) + iy}}{{x + i\left( {y + 1} \right)}}} \right) = 0\\\ \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\left( {\dfrac{{\left( {x + 1} \right) + iy}}{{x + i\left( {y + 1} \right)}}} \right) \times \left( {\dfrac{{x - i\left( {y + 1} \right)}}{{x - i\left( {y + 1} \right)}}} \right)} \right] = 0\\\ \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\dfrac{{\left\\{ {(x + 1) + iy} \right\\}\left\\{ {x - i(y + 1)} \right\\}}}{{{x^2} + {{(y + 1)}^2}}}} \right] = 0\\\ \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\dfrac{{x(x + 1) + y(y + 1) + i\left\\{ {xy - (x + 1)(y + 1)} \right\\}}}{{{x^2} + {{(y + 1)}^2}}}} \right] = 0\\\ \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\dfrac{{x(x + 1) + y(y + 1)}}{{{x^2} + {{(y + 1)}^2}}} + i\dfrac{{xy - (x + 1)(y + 1)}}{{{x^2} + {{(y + 1)}^2}}}} \right] = 0 \end{array}$$ Clearly the portion not containing i is the real part $$\begin{array}{l} \therefore \dfrac{{x(x + 1) + y(y + 1)}}{{{x^2} + {{(y + 1)}^2}}} = 0\\\ \Rightarrow x(x + 1) + y(y + 1) = 0\\\ \Rightarrow {x^2} + x + {y^2} + y = 0 \end{array}$$ By using the process of completing squares this becomes $$\begin{array}{l} \Rightarrow {x^2} + 2 \times \dfrac{1}{2} \times x + {\left( {\dfrac{1}{2}} \right)^2} + {y^2} + 2 \times y \times \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^2} + {\left( {\dfrac{1}{2}} \right)^2}\\\ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {y + \dfrac{1}{2}} \right)^2} = \dfrac{1}{4} + \dfrac{1}{4}\\\ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {y + \dfrac{1}{2}} \right)^2} = \dfrac{2}{4}\\\ \Rightarrow {\left( {x + \dfrac{1}{2}} \right)^2} + {\left( {y + \dfrac{1}{2}} \right)^2} = \dfrac{1}{2} \end{array}$$ Now we know that the general equation of a circle is given by $${\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}$$ where (h,k) is the center of the circle and r is the radius of the circle. Now comparing this with the equation we have The center is $$\left( { - \dfrac{1}{2}, - \dfrac{1}{2}} \right)$$ and the radius is $$\begin{array}{l} {r^2} = \dfrac{1}{2}\\\ r = \sqrt {\dfrac{1}{2}} \end{array}$$ Clearly it is a circle with $$radius \ne 1$$ and for option D let us check by putting (1,1) in the equation of circle $$\begin{array}{l} {x^2} + {y^2} + x + y\\\ = {1^2} + {1^2} + 1 + 1\\\ = 4 \end{array}$$ And clearly $$4 \ne 0$$ therefore option d is also incorrect Therefore option B is the correct option. Note: As it was having a quadratic equation therefore it was clear that it is not a straight line as a straight line contains a linear equation and also Re was denoted for the real part in the complex number.