Question

Consider the real part of $\dfrac{\left( z+4 \right)}{\left( 2z+i \right)}$ is equal to $\dfrac{1}{2}$ , where $i=\sqrt{-2}$ , then the point z lies on a ……….

Answer 1

Hint: Substitute z as any complex number, say a + ib and find relation between a, b then say about its locus of a, b is nothing but the point z’s locus.


Complete step-by-step solution -
Definition of i, can be written as:
The solution of the equation: ${{x}^{2}}+1=0$ is i. i is an imaginary number. Any number which has an imaginary number in its representation is called a complex number.
Definition of a complex equation, can be written as: An equation containing complex numbers in it, is called a complex equation.
It is possible to have a real root for complex equations.
Example: $\left( 1+i \right)x+\left( 1+i \right)=0,x= -1$ is the root of the equation.
Let us assume:
$z=a+ib$
We need a real part of an expression so for easy solving we are taking this assumption.
By substituting the z value into the equation, we get:
Given equation in the question on the term z is:
$\operatorname{Re}\left[ \dfrac{z+4}{2z-i} \right]=\dfrac{1}{2}$
By using the substitution of z into the equation, we get:
$\begin{align}
  & \operatorname{Re}\left[ \dfrac{\left( x+iy \right)+4}{2\left( x+iy \right)-i} \right]=\dfrac{1}{2} \\
 & \Rightarrow \operatorname{Re}\left[ \dfrac{x+iy+4}{2x+2iy-i} \right]=\dfrac{1}{2} \\
\end{align}$
By simplifying the equation of $x,y,$ we make it as:
By grouping real part and imaginary part together, we get:
\[\operatorname{Re}\left[ \dfrac{\left( x+4 \right)+\left( iy \right)}{2x+i\left( 2y-1 \right)} \right]=\dfrac{1}{2}\]
By rationalizing the term inside the bracket, we turn into:
By multiplying and dividing with \[\left( 2x-i\left( 2y-1 \right) \right)\] term, we turn into:
$\operatorname{Re}\left[ \dfrac{\left( \left( x+4 \right)+\left( iy \right) \right)\left( 2x-i\left( 2y-1 \right) \right)}{\left( 2x+i\left( 2y-1 \right) \right)\left( 2x-i\left( 2y-1 \right) \right)} \right]=\dfrac{1}{2}$
By simplifying this equation of $x,y,$ we turn this into:
$\dfrac{2{{x}^{2}}+5x+2{{y}^{2}}-y}{4{{x}^{2}}+4{{y}^{2}}-4y+1}=\dfrac{1}{2}$
By cross multiplication of both terms $x,y,$we turn this into:
$4{{x}^{2}}+10x+4{{y}^{2}}-2y=4{{x}^{2}}+4{{y}^{2}}-4y+1$
By cancelling the common second order terms, we turn this into:
$10x+2y=1$
Therefore, z lies on a straight line with equation $10x+2y=1$

Note: Be careful while solving the real part after rationalization as the number of terms is large. Here we need to be careful while multiplying two complex number as only we need to find real part so multiply real part to real part and imaginary part to another number’s imaginary part as we know value of $i = \sqrt{-1} $ so, value of ${i^2} = -1 $,hence imaginary part becomes real.