Question

If complex numbers ${{z}_{1}}$, ${{z}_{2}}$ and ${{z}_{3}}$ represent the vertices $A$, $B$ and $C$ respectively of an isosceles triangle $ABC$ of which $\angle C$ is right angle, then the correct statement is
A. ${{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{3}}^{2}={{z}_{1}}{{z}_{2}}{{z}_{3}}$
B. ${{({{z}_{3}}-{{z}_{1}})}^{2}}={{z}_{3}}-{{z}_{2}}$
C. ${{({{z}_{1}}-{{z}_{2}})}^{2}}=({{z}_{1}}-{{z}_{3}})({{z}_{3}}-{{z}_{2}})$
D. ${{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})({{z}_{3}}-{{z}_{2}})$

Answer 1

Hint: In this question, we have to find the relationship between the given complex numbers. To find this, the properties of a triangle that are the given triangle is an isosceles-right angle triangle are used.

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, $\cos \theta =\frac{x}{r},\sin \theta =\frac{b}{r}$
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 triangle has vertices
$A$ represented by the complex number ${{z}_{1}}$,
$B$ represented by the complex number ${{z}_{2}}$, and
$C$ represented by the complex number ${{z}_{3}}$
It is given that, the triangle $ABC$ is an isosceles triangle. That means any two sides are equal in length. I.e.,
$BC=CA\text{ }...(1)$
The triangle has right angle at $C$. So, $\angle C=\frac{\pi }{2}$.
Since it is a right-angle triangle, we can apply the Pythagoras theorem. I.e.,
$B{{A}^{2}}=B{{C}^{2}}+C{{A}^{2}}\text{ }...(2)$
Substituting (1) in (2)
$\begin{align}
  & B{{A}^{2}}=B{{C}^{2}}+B{{C}^{2}}\text{ } \\
 & \Rightarrow B{{A}^{2}}=2B{{C}^{2}}\text{ }...(3) \\
\end{align}$
Thus, substituting their complex values in (3), we get
${{({{z}_{1}}-{{z}_{2}})}^{2}}=2{{({{z}_{3}}-{{z}_{2}})}^{2}}\text{ }...(4)$
But we can write
$\begin{align}
  & BC=CA \\
 & \Rightarrow ({{z}_{3}}-{{z}_{2}})=({{z}_{1}}-{{z}_{3}}) \\
\end{align}$
Applying this in (4), we get
$\begin{align}
  & {{({{z}_{1}}-{{z}_{2}})}^{2}}=2{{({{z}_{3}}-{{z}_{2}})}^{2}} \\
 & \Rightarrow {{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{3}}-{{z}_{2}})({{z}_{3}}-{{z}_{2}}) \\
 & \Rightarrow {{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{3}}-{{z}_{2}})({{z}_{1}}-{{z}_{3}}) \\
\end{align}$
Thus, the correct statement is ${{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})({{z}_{3}}-{{z}_{2}})$.

Option ‘D’ is correct

Note: Here, we have to apply the Pythagoras theorem to get the required statement. We can also calculate this by rotating the vertex $C$ in anticlockwise, so that we can write $CB=CA\cdot {{e}^{i\frac{\pi }{2}}}$. On evaluating this, we get the required statement.