Question

If ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ are three collinear points in argand plane, then \[\left| \begin{matrix}
   {{z}_{1}} & \overline{{{z}_{1}}} & 1 \\
   {{z}_{2}} & \overline{{{z}_{2}}} & 1 \\
   {{z}_{3}} & \overline{{{z}_{3}}} & 1 \\
\end{matrix} \right|=\]
A. $0$
B. $-1$
C. $1$
D. $2$

Answer 1

Hint: In this question, we are to find the determinant of the given matrix. Here the matrix consists complex number and their conjugates. Since the given points are collinear, the mod amplitudes of the given complex numbers are equal. So, by this we can calculate the determinant of the given matrix.



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 $\cos \theta +i\sin \theta ={{e}^{i\theta }}$



Complete step by step solution:Given that, ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ are three collinear points. So, the mod amplitudes of these collinear points are equal. I.e., $\left| {{z}_{1}} \right|=\left| {{z}_{2}} \right|=\left| {{z}_{3}} \right|=k$
We have
$z=x+iy$ and $\overline{z}=x-iy$
Then,
$\begin{align}
  & \left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
 & \left| \overline{z} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}} \\
 & \Rightarrow \left| z \right|=\left| \overline{z} \right| \\
\end{align}$
From this, we can write
$\begin{align}
  & \left| {{z}_{1}} \right|=\left| \overline{{{z}_{1}}} \right|=k \\
 & \left| {{z}_{2}} \right|=\left| \overline{{{z}_{2}}} \right|=k \\
 & \left| {{z}_{3}} \right|=\left| \overline{{{z}_{3}}} \right|=k \\
\end{align}$
So, on substituting these values in the matrix, we get the determinant as zero.
This is because, if the two rows of a matrix are the same then its determinant is zero.
Therefore,
\[\left| \begin{matrix}
   {{z}_{1}} & \overline{{{z}_{1}}} & 1 \\
   {{z}_{2}} & \overline{{{z}_{2}}} & 1 \\
   {{z}_{3}} & \overline{{{z}_{3}}} & 1 \\
\end{matrix} \right|=0\]



Option ‘A’ is correct



Note: Here we need to remember that, if the two rows of a matrix are the same then the determinant of that matrix is zero. With this statement, we can conclude the given matrix.