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
\[\begin{array}{l}
\therefore {\mathop{\rm Re}\nolimits} \left( {\dfrac{{\left( {x + 1} \right) + iy}}{{x + i\left( {y + 1} \right)}}} \right) = 0\\
 \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\left( {\dfrac{{\left( {x + 1} \right) + iy}}{{x + i\left( {y + 1} \right)}}} \right) \times \left( {\dfrac{{x - i\left( {y + 1} \right)}}{{x - i\left( {y + 1} \right)}}} \right)} \right] = 0\\
 \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\dfrac{{\left\{ {(x + 1) + iy} \right\}\left\{ {x - i(y + 1)} \right\}}}{{{x^2} + {{(y + 1)}^2}}}} \right] = 0\\
 \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\dfrac{{x(x + 1) + y(y + 1) + i\left\{ {xy - (x + 1)(y + 1)} \right\}}}{{{x^2} + {{(y + 1)}^2}}}} \right] = 0\\
 \Rightarrow {\mathop{\rm Re}\nolimits} \left[ {\dfrac{{x(x + 1) + y(y + 1)}}{{{x^2} + {{(y + 1)}^2}}} + i\dfrac{{xy - (x + 1)(y + 1)}}{{{x^2} + {{(y + 1)}^2}}}} \right] = 0
\end{array}\]
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.