Question

If $ img \left( \dfrac{z-1}{2z+1} \right)=-4 $ , then the locus of $ z $ is
A. an ellipse
B. a parabola
C. a straight line
D. a circle

Answer 1

Hint: We first assume the complex number $ z=a+ib $ . We find the simplified form of the expression $ \dfrac{z-1}{2z+1} $ . We rationalise it and find the argument of the complex form. We take the equation and equate it with the general form of circle to find the solution.

Complete step-by-step answer:
Let us assume the complex number $ z=a+ib $ where $ a,b\in \mathbb{R} $ . The argument for $ z=a+ib $ can be denoted as $ {{\tan }^{-1}}\left( \dfrac{b}{a} \right) $ .
The function $ \dfrac{z-1}{2z+1} $ becomes $ \dfrac{z-1}{2z+1}=\dfrac{a+ib-1}{2\left( a+ib \right)+1}=\dfrac{\left( a-1 \right)+ib}{\left( 2a+1 \right)+2ib} $ .
We first apply the rationalisation of complex numbers for $ \dfrac{\left( a-1 \right)+ib}{\left( 2a+1 \right)+2ib} $ .
We multiply $ \left( 2a+1 \right)-2ib $ to both the numerator and denominator of the fraction.
So, \[\dfrac{\left( a-1 \right)+ib}{\left( 2a+1 \right)+2ib}\times \dfrac{\left( 2a+1 \right)-2ib}{\left( 2a+1 \right)-2ib}=\dfrac{\left( a-1 \right)\left( 2a+1 \right)-2{{i}^{2}}{{b}^{2}}-ib\left( 2a-2-2a-1 \right)}{{{\left( 2a+1 \right)}^{2}}-4{{i}^{2}}{{b}^{2}}}\] .
We used the theorem of \[\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}\] .
 \[\begin{align}
  & \dfrac{\left( a-1 \right)+ib}{\left( 2a+1 \right)+2ib} \\
 & =\dfrac{\left( a-1 \right)\left( 2a+1 \right)-2{{i}^{2}}{{b}^{2}}-ib\left( 2a-2-2a-1 \right)}{{{\left( 2a+1 \right)}^{2}}-4{{i}^{2}}{{b}^{2}}} \\
 & =\dfrac{\left( a-1 \right)\left( 2a+1 \right)+2{{b}^{2}}}{4{{a}^{2}}+4a+1+4{{b}^{2}}}+i\dfrac{3b}{4{{a}^{2}}+4a+1+4{{b}^{2}}} \\
\end{align}\]
Therefore,
 \[\begin{align}
  & img \left( \dfrac{z-1}{2z+1} \right)=-4 \\
 & \Rightarrow img \left( \dfrac{\left( a-1 \right)\left( 2a+1 \right)+2{{b}^{2}}}{4{{a}^{2}}+4a+1+4{{b}^{2}}}+i\dfrac{3b}{4{{a}^{2}}+4a+1+4{{b}^{2}}} \right)=-4 \\
 & \Rightarrow \dfrac{3b}{4{{a}^{2}}+4a+1+4{{b}^{2}}}=-4 \\
\end{align}\]
This gives
 \[\begin{align}
  & \dfrac{3b}{4{{a}^{2}}+4a+1+4{{b}^{2}}}=-4 \\
 & \Rightarrow 16{{a}^{2}}+16a+4+16{{b}^{2}}+3b=0 \\
\end{align}\]
This is an equation of circle of
 \[\begin{align}
  & 16{{a}^{2}}+16a+4+16{{b}^{2}}+3b=0 \\
 & \Rightarrow 4{{\left( 2a+1 \right)}^{2}}+{{\left( 4b+\dfrac{3}{8} \right)}^{2}}={{\left( \dfrac{3}{8} \right)}^{2}} \\
 & \Rightarrow {{\left( a+\dfrac{1}{2} \right)}^{2}}+{{\left( b+\dfrac{3}{32} \right)}^{2}}={{\left( \dfrac{3}{32} \right)}^{2}} \\
\end{align}\] .
The circle is with centre $ \left( -\dfrac{1}{2},-\dfrac{3}{32} \right) $ and radius $ \dfrac{3}{32} $ as the general equation of circle is \[{{\left( x-m \right)}^{2}}+{{\left( y-n \right)}^{2}}={{r}^{2}}\] having centre $ \left( m,n \right) $ and radius $ r $ .
The correct option is D.
So, the correct answer is “Option D”.

Note: The integer power value of every even number on $ i $ will give the real number. Odd numbers of power value give back the imaginary part. The relation $ {{i}^{2}}=-1,{{i}^{3}}=-i,{{i}^{4}}=1 $ can also be represented as the unit circle on the complex plane.