Question

If $z$ is a complex number in the argand plane, then the equation $\left| z-2 \right|+\left| z+2 \right|=8$ represents
A. Parabola
B. Ellipse
C. Hyperbola
D. Circle

Answer 1

Hint: In this question, we are to find the type of the given equation. The given options are the curves, by using the eccentricity of the curve. Depending on the eccentricity, the given equation is named.

Formula used: The complex number $(x,y)$ is represented by $x+iy$.
If $z=x+iy\in C$, then $x$ is called the real part and $y$ is called the imaginary part of $z$. These are represented by $\operatorname{Re}(z)$ and $\operatorname{Im}(z)$ respectively.
$z=x+iy$ be a complex number such that $\left| z \right|=r$ and $\theta $ be the amplitude of $z$. So, $\cos \theta =\dfrac{x}{r},\sin \theta =\dfrac{b}{r}$
And we can write the magnitude as
 $\begin{align}
  & \left| z \right|=\left| x+iy \right| \\
 & \Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
\end{align}$
Thus, we can write
$z=x+iy=r\cos \theta +ir\sin \theta =r(\cos \theta +i\sin \theta )$
This is said to be the mod amplitude form or the polar form of $z$.
Where $\cos \theta +i\sin \theta $ is denoted by $cis\theta $ and the Euler’s formula is $\cos \theta +i\sin \theta ={{e}^{i\theta }}$
The eccentricity is calculated by the intercepts of the axes,
$e=\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a}$
If the eccentricity $e=1$, then the curve is a parabola
If the eccentricity $e<1$, then the curve is an ellipse
If the eccentricity $e>1$, then the curve is a hyperbola

Complete step by step solution: The given equation is $\left| z-2 \right|+\left| z+2 \right|=8$
We know that, $z=x+iy$
On substituting,
\[\begin{align}
  & \left| z-2 \right|+\left| z+2 \right|=8 \\
 & \left| x+iy-2 \right|+\left| x+iy+2 \right|=8 \\
 & \left| (x-2)+iy \right|+\left| (x+2)+iy \right|=8 \\
 & \sqrt{{{(x-2)}^{2}}+{{y}^{2}}}+\sqrt{{{(x+2)}^{2}}+{{y}^{2}}}=8 \\
 & \sqrt{{{(x-2)}^{2}}+{{y}^{2}}}=8-\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
\end{align}\]
Squaring on both sides, we get
\[\begin{align}
  & {{(x-2)}^{2}}+{{y}^{2}}=64+{{(x+2)}^{2}}+{{y}^{2}}-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
 & {{x}^{2}}-4x+4+{{y}^{2}}=64+{{x}^{2}}+4x+4+{{y}^{2}}-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
 & -4x-4x=64-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
 & -8x-64=-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
\end{align}\]
\[\begin{align}
  & -8(x+8)=-16\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
 & (x+8)=2\sqrt{{{(x+2)}^{2}}+{{y}^{2}}} \\
 & {{(x+8)}^{2}}=4{{(x+2)}^{2}}+4{{y}^{2}} \\
 & {{x}^{2}}+16x+64=4({{x}^{2}}+4x+4)+4{{y}^{2}} \\
\end{align}\]
\[\begin{align}
  & {{x}^{2}}+16x+64=4{{x}^{2}}+16x+16+4{{y}^{2}} \\
 & 3{{x}^{2}}+4{{y}^{2}}=48 \\
 & \dfrac{3{{x}^{2}}}{48}+\dfrac{4{{y}^{2}}}{48}=1 \\
 & \dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{12}=1 \\
\end{align}\]
Here the intercepts are $a=16,b=12$.
Since $a>b$, the eccentricity is
$\begin{align}
  & e=\dfrac{\sqrt{{{a}^{2}}-{{b}^{2}}}}{a} \\
 & \text{ }=\dfrac{\sqrt{{{(16)}^{2}}-{{(12)}^{2}}}}{16} \\
 & \text{ }=\dfrac{\sqrt{112}}{16}<1 \\
\end{align}$
Since the eccentricity $e<1$, the given equation represents an ellipse.

Thus, Option (B) is correct.

Note: Here we need to remember the eccentricity is to be calculated in order to find the type of the curve.