 QUESTION

# If $z_1$, $z_2$, $z_3$ are unimodular complex numbers then the greatest value of $|{{z}_{1}}-{{z}_{2}}{{|}^{2}}+|{{z}_{2}}-{{z}_{3}}{{|}^{2}}+|{{z}_{3}}-{{z}_{1}}{{|}^{2}}$ is.(a) 3(b) 6(c) 9(d) $\dfrac{27}{2}$

Hint: In this question, we first need to expand each term and then substitute the unimodular condition to simplify the expanded terms. Then use the inequality that the modulus of the three complex numbers should be always greater than zero which gives us the final result.

COMPLEX NUMBER: A number of the form $z=x+iy$, where $x,y\in R$, is called a complex number.
Where x is the real part and y is the imaginary part of the given complex number.
CONJUGATE OF A COMPLEX NUMBER:
If $z=x+iy$ is a complex number, then the conjugate of this complex number z is denoted by $\bar{z}$
$\bar{z}=x-iy$
MODULUS OF A COMPLEX NUMBER: If $z=x+iy$, then modulus or magnitude of z is denoted by $|z|$ and is given by
$|z|=\sqrt{{{x}^{2}}+{{y}^{2}}}$
Z is unimodular , if $|z|=1$
Now, from the given equation in the question,
$|{{z}_{1}}-{{z}_{2}}{{|}^{2}}+|{{z}_{2}}-{{z}_{3}}{{|}^{2}}+|{{z}_{3}}-{{z}_{1}}{{|}^{2}}$
Let us assume this function as some E
$\Rightarrow E=|{{z}_{1}}-{{z}_{2}}{{|}^{2}}+|{{z}_{2}}-{{z}_{3}}{{|}^{2}}+|{{z}_{3}}-{{z}_{1}}{{|}^{2}}$
Now, let us expand each term in the above equation separately
$\Rightarrow |{{z}_{1}}-{{z}_{2}}{{|}^{2}}=\left( {{z}_{1}}-{{z}_{2}} \right)\left( {{{\bar{z}}}_{1}}-{{{\bar{z}}}_{2}} \right)$
Now, this can be further simplified as.
$\Rightarrow |{{z}_{1}}-{{z}_{2}}{{|}^{2}}=|{{z}_{1}}{{|}^{2}}+|{{z}_{2}}{{|}^{2}}-\left( {{z}_{1}}{{{\bar{z}}}_{2}}+{{{\bar{z}}}_{1}}{{z}_{2}} \right)$
As we already know the modulus of the functions this can be further written as.
\begin{align} & \Rightarrow |{{z}_{1}}-{{z}_{2}}{{|}^{2}}=1+1-\left( {{z}_{1}}{{{\bar{z}}}_{2}}+{{{\bar{z}}}_{1}}{{z}_{2}} \right) \\ & \Rightarrow |{{z}_{1}}-{{z}_{2}}{{|}^{2}}=2-2\operatorname{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right) \\ \end{align}
Similarly, the other term can also be expanded as.
$\Rightarrow |{{z}_{2}}-{{z}_{3}}{{|}^{2}}=\left( {{z}_{2}}-{{z}_{3}} \right)\left( {{{\bar{z}}}_{2}}-{{{\bar{z}}}_{3}} \right)$
Now, this can be further simplified as.
$\Rightarrow |{{z}_{2}}-{{z}_{3}}{{|}^{2}}=|{{z}_{2}}{{|}^{2}}+|{{z}_{3}}{{|}^{2}}-\left( {{z}_{2}}{{{\bar{z}}}_{3}}+{{{\bar{z}}}_{2}}{{z}_{3}} \right)$
As we already know the modulus of the functions this can be further written as.
\begin{align} & \Rightarrow |{{z}_{2}}-{{z}_{3}}{{|}^{2}}=1+1-\left( {{z}_{2}}{{{\bar{z}}}_{3}}+{{{\bar{z}}}_{2}}{{z}_{3}} \right) \\ & \Rightarrow |{{z}_{2}}-{{z}_{3}}{{|}^{2}}=2-2\operatorname{Re}\left( {{z}_{2}}{{{\bar{z}}}_{3}} \right) \\ \end{align}
Similarly, again the third term can be expanded as.
$\Rightarrow |{{z}_{3}}-{{z}_{1}}{{|}^{2}}=\left( {{z}_{3}}-{{z}_{1}} \right)\left( {{{\bar{z}}}_{3}}-{{{\bar{z}}}_{1}} \right)$
Now, this can be further simplified as.
$\Rightarrow |{{z}_{3}}-{{z}_{1}}{{|}^{2}}=|{{z}_{3}}{{|}^{2}}+|{{z}_{1}}{{|}^{2}}-\left( {{z}_{3}}{{{\bar{z}}}_{1}}+{{{\bar{z}}}_{3}}{{z}_{1}} \right)$
As we already know the modulus of the functions this can be further written as.
\begin{align} & \Rightarrow |{{z}_{3}}-{{z}_{1}}{{|}^{2}}=1+1-\left( {{z}_{3}}{{{\bar{z}}}_{1}}+{{{\bar{z}}}_{3}}{{z}_{1}} \right) \\ & \Rightarrow |{{z}_{3}}-{{z}_{1}}{{|}^{2}}=2-2\operatorname{Re}\left( {{z}_{3}}{{{\bar{z}}}_{1}} \right) \\ \end{align}
Now, the complete equation can be written as.
$\Rightarrow E=2-2\operatorname{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}} \right)+2-2\operatorname{Re}\left( {{z}_{2}}{{{\bar{z}}}_{3}} \right)+2-2\operatorname{Re}\left( {{z}_{3}}{{{\bar{z}}}_{1}} \right)$
This can be further simplified as.
\begin{align} & \Rightarrow E=2+2+2-2\operatorname{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}}+{{z}_{2}}{{{\bar{z}}}_{3}}+{{z}_{3}}{{{\bar{z}}}_{1}} \right) \\ & \Rightarrow E=6-2\operatorname{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}}+{{z}_{2}}{{{\bar{z}}}_{3}}+{{z}_{3}}{{{\bar{z}}}_{1}} \right) \\ \end{align}
Now, as we already know that the modulus function is greater than 0 we can write.
$\Rightarrow |{{z}_{1}}+{{z}_{2}}+{{z}_{3}}{{|}^{2}}\ge 0$
Now, on expanding this equation accordingly we get,
$\Rightarrow 1+1+1+2\operatorname{Re}\left( {{z}_{1}}{{{\bar{z}}}_{2}}+{{z}_{2}}{{{\bar{z}}}_{3}}+{{z}_{3}}{{{\bar{z}}}_{1}} \right)\ge 0$
Now, on substituting the value of the real part from the above relation between E and real part we get,
\begin{align} & \Rightarrow 3+\left( 6-E \right)\ge 0 \\ & \Rightarrow 9-E\ge 0 \\ \end{align}
Now, on rearranging the terms we get,
$\therefore E\le 9$
Hence, the correct option is (c).

Note: It is important to note that the modulus of a function is greater than or equals to zero because it includes the squaring of the terms which cannot be negative. By applying that condition only we can get the highest possible value of the given function.
While rearranging the terms we need to be careful about the signs and the terms because neglecting any one of them changes the result completely. It is also important to note that when we take the terms to the other side then the inequality changes accordingly because taking the opposite sign changes the result.