Question

Let $z$ be the set of complex numbers. Then the equation ,$2\left| z+3i \right|-\left| z-i \right|=0$ represents\[\]
A. A circle with radius $\dfrac{10}{3}$\[\]
B. A circle with radius $\dfrac{8}{3}$\[\]
C. An ellipse with the length of minor axis $\dfrac{16}{9}$\[\]
D. An ellipse with length of major axis $\dfrac{16}{3}$\[\]

Answer 1

Hint: Substitute $z=x+iy$ in the given equation and simplify to get an equation involving $x$and $y$.Compare the obtained equation with standard second degree equation of any circle and ellipse in a plane (as there are two variables) to find out the correct option(s).\[\]

Complete step by step answer:
The given equation is
\[2\left| z+3i \right|-\left| z-i \right|=0...(1)\]
We put $z=x+iy$ in above equation where $x$ and $y$ are any real numbers.
\[\begin{align}
  & 2\left| \left( x+iy \right)+3i \right|-\left| \left( x+iy \right)-i \right|=0 \\
 & \Rightarrow 2\left| x+i\left( 3+y \right) \right|-\left| x-i\left( y-1 \right) \right|=0 \\
\end{align}\]
We know that the modulus of a complex x number is defined as \[\left| z \right|=\left| x\pm iy \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}\]. Using it to advance,
\[\begin{align}
  & \Rightarrow 2\left| x+i\left( 3+y \right) \right|-\left| x-i\left( y-1 \right) \right|=0 \\
 & \Rightarrow 2\sqrt{{{x}^{2}}+{{\left( 3+y \right)}^{2}}}=\sqrt{{{x}^{2}}+{{\left( y-1 \right)}^{2}}} \\
\end{align}\]
Squaring both side,
\[\begin{align}
  & \Rightarrow 4\left( {{x}^{2}}+{{\left( 3+y \right)}^{2}} \right)={{x}^{2}}+{{\left( y-1 \right)}^{2}} \\
 & \Rightarrow 4{{x}^{2}}+4\left( {{y}^{2}}+6y+9 \right)-{{x}^{2}}-\left( {{y}^{2}}-2y+1 \right)=0 \\
 & \Rightarrow 3{{x}^{2}}+3{{y}^{2}}+26y+35=0 \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}+\dfrac{26}{3}y+\dfrac{35}{3}=0....(2) \\
\end{align}\]
We know from the general second degree equation in two variables with real constants $a,b,g,f,c$ is given by
\[a{{x}^{2}}+b{{y}^{2}}+2hxy+2gx+2fy+c=0\]
 The general second degree equation is an equation of the circle in a plane if ${{h}^{2}}-ab<0$ , coefficient of ${{x}^{2}}$ is same as coefficient of ${{y}^{2}}$ , $ a=b $ and the value of $h=0$. The general equation of circle is
\[{{x}^{2}}+{{y}^{2}}+2gx+2fy+c=0\]
Comparing it with equation (2) we observe that equation (2) is an equation of a circle. The radius of a circle with the standard form is given by $\sqrt{{{g}^{2}}+{{f}^{2}}-c}$. We get from equation(2) that radius of the circle is $\sqrt{{{g}^{2}}+{{f}^{2}}-c}=\sqrt{{{0}^{2}}+{{\left( \dfrac{1}{2}\times \dfrac{26}{9} \right)}^{2}}-\dfrac{35}{3}}=\sqrt{\dfrac{64}{9}}=\dfrac{8}{3}$.\[\]
The general second degree equation in two variables is an equation of ellipse when ${{h}^{2}}-ab<0$ and coefficient of ${{x}^{2}}$ is NOT same as coefficient of ${{y}^{2}}$$\left( a\ne b \right)$ which is not true in case of equation(2). So equation (2) is not an equation of an ellipse.

So, the correct answer is “Option B”.

Note: The question tests your knowledge of modulus of complex numbers and the general equation of circle and ellipse in two dimensions. Careful solving of simultaneous equation substitution and usage of formula will lead us to arrive at the correct result. You can also solve for equations involving more than one complex variable.