Question

If ${{z}_{1}},{{z}_{2}},{{z}_{3}}$ are affixes of the vertices $A,B,C$ respectively of a triangle $ABC$ having centroid at $G$ such that $z=0$ is the midpoint of $AG$, then
A. ${{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0$
B. ${{z}_{1}}+4{{z}_{2}}+{{z}_{3}}=0$
C. ${{z}_{1}}+{{z}_{2}}+4{{z}_{3}}=0$
D. $4{{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0$

Answer 1

Hint: In this question, we are to find the equation for the midpoint of the line $AG$, in the given triangle $ABC$ having centroid at $G$. For this, we have the affix of the vertex $A$ and the affix for the midpoint of $AG$. So, we can find the required equation.

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 affixes of the vertices $A,B,C$ respectively of a triangle $ABC$ having centroid at $G$.
It is also given that the affix of the midpoint of $AG$ as $z=0$.
But we know that,
Centroid \[G=\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3}\]
We have the affixes for the vertex $A$ and $G$. So, their midpoint is
$\begin{align}
  & =\dfrac{\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3}+{{z}_{1}}}{2} \\
 & =\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+3{{z}_{1}}}{6} \\
 & =\dfrac{4{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{6} \\
\end{align}$
But it is given that, the affix of the midpoint is $z=0$
Thus,
$\begin{align}
  & \dfrac{4{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{6}=0 \\
 & 4{{z}_{1}}+{{z}_{2}}+{{z}_{3}}=0 \\
\end{align}$

Thus, Option (D) is correct.

Note: By applying the appropriate formulae, we can calculate the required equation. Here we need to remember that the centroid is the ratio of the sum of the affixes of the vertices of the triangle to the number of vertices of the triangle.