Question

The Locus of Z satisfying the equation, \[\left| \dfrac{\left( z+2i \right)}{\left( 2z+i \right)} \right|<1\], where z = x + iy is:
A. \[{{x}^{2}}+{{y}^{2}}<1\]
B. \[{{x}^{2}}-{{y}^{2}}<1\]
C. \[{{x}^{2}}+{{y}^{2}}>1\]
D. \[2{{x}^{2}}+3{{y}^{2}}<1\]

Answer 1

Hint: First we are going to substitute the value of z in the equation that has been mentioned in the question and then solve the complex number by multiplying the denominator that is present on the left hand side (LHS) of the inequality equation to the right hand side (RHS) of the inequality equation

Complete step by step solution:
As stated in the question we know the value of z = x + iy,
We will start by substituting the value of z = x+iy in the main inequality equation and we get\[\begin{align}
  & \Rightarrow \left| \dfrac{\left( z+2i \right)}{\left( 2z+i \right)} \right|<1 \\
 & \Rightarrow \left| \dfrac{\left( x+iy+2i \right)}{\left( 2\left( x+iy \right)+i \right)} \right|<1 \\
 & \Rightarrow \left| \dfrac{\left( x+i\left( y+2 \right) \right)}{\left( 2x+i\left( 2y+1 \right) \right)} \right|<1 \\
 & \Rightarrow \left| \left( x+i\left( y+2 \right) \right) \right|<\left| \left( \left( 2x+i\left( 2y+1 \right) \right) \right) \right|............(1) \\
\end{align}\]
In the above solution we can see that after substituting we got equation 1 which is still in the complex form. Now to change the complex from into equation we know that \[{{z}^{2}}={{x}^{2}}+{{y}^{2}}\] through which we got the equation (2).
\[\Rightarrow \sqrt{\left( {{x}^{2}}+{{\left( y+2 \right)}^{2}} \right)}<\sqrt{{{\left( 2x \right)}^{2}}+{{\left( 2y+1 \right)}^{2}}}\text{ }\left[ \text{now we will be squaring both sides} \right]............\left( 2 \right)\]
After we changed it to equation form, we will then square both sides of the inequality with which we will get the end result as equation 3,
\[\Rightarrow \left( {{x}^{2}}+{{y}^{2}}+4+4y \right)<\left( 4{{x}^{2}}+4{{y}^{2}}+1+4y \right).........\left( 3 \right)\]
Now we can see that the RHS of the equation has higher value of \[{{x}^{2}}\] and \[{{y}^{2}}\] i.e. coefficients of \[{{x}^{2}}\] and \[{{y}^{2}}\],so we will shift the variables on the Right hand side of the inequality and take the constant to the left hand side of the inequality for which we will get a new equation which is equation 4.
\[\Rightarrow 3{{x}^{2}}+3{{y}^{2}}>3........\left( 4 \right)\]
In equation 3 we will then take 3 out on the left hand side of the inequality, which will then be cancelled out by 3 on the right hand side of the equation as both of them are multiple of 3, and then we will be getting our final answer i.e. the locus of Z.
\[\Rightarrow {{x}^{2}}+{{y}^{2}}>1........\left( 5 \right)\]
So, for the question mentioned, the locus of Z is option (c) i.e. \[{{x}^{2}}+{{y}^{2}}>1\]

Note: In this remember to change the complex form of the equation to normal variable equation ( this will help us to solve the equations easily). To change the equation in normal form we know the relation that \[\left| z \right|=\left| {{x}^{2}}+{{y}^{2}} \right|\] where the complex equation for z = x+iy.