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 isA. ${{z}_{1}}^{2}+{{z}_{2}}^{2}+{{z}_{3}}^{2}={{z}_{1}}{{z}_{2}}{{z}_{3}}$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. ${{({{z}_{1}}-{{z}_{2}})}^{2}}=2({{z}_{1}}-{{z}_{3}})({{z}_{3}}-{{z}_{2}})$

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Vedantu Content Team · Accepted Answer

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 correctNote: 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.

If complex numbers z1, z2 and z3 represent the vertices A, B and C respectively of an isosceles triangle ABC of which ∠C is right angle, then the correct statement isA. z12+z22+z32=z1z2z3B. (z3−z1)2=z3−z2C. (z1−z2)2=(z1−z3)(z3−z2)D. (z1−z2)2=2(z1−z3)(z3−z2)