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 $\left( a+b \right)\times \left( a-b \right)={{a}^{2}}-{{b}^{2}}$ .

& \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{aligned}$$ Therefore, $$\begin{aligned} & 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{aligned}$$ This gives $$\begin{aligned} & \dfrac{3b}{4{{a}^{2}}+4a+1+4{{b}^{2}}}=-4 \\\ & \Rightarrow 16{{a}^{2}}+16a+4+16{{b}^{2}}+3b=0 \\\ \end{aligned}$$ This is an equation of circle of $$\begin{aligned} & 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{aligned}$$ . The circle is with centre $ \left( -\dfrac{1}{2},-\dfrac{3}{32} \right) $ and radius $ \dfrac{3}{32} $ as the general equation of circle is $${{\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.