Show that the center of mass of three identical particles situated at the point ${\mathit{z}}_{\mathbf{1}}\mathbf{,}{\mathit{z}}_{\mathbf{2}}\mathbf{,}{\mathit{z}}_{\mathbf{3}}$is $\frac{\left({\mathit{z}}_{\mathbf{1}}\mathbf{,}{\mathit{z}}_{\mathbf{2}}\mathbf{,}{\mathit{z}}_{\mathbf{3}}\right)}{\mathbf{3}}$.

To prove thatthe center of mass of three identical particles situated at the point$z_1,z_2,z_3$is $\frac{\left(z_1,z_2,z_3\right)}{3}$.

Complex numbers possess real numbers and imaginary numbers; a complex can be written in the form of:

z = x +iy

Here x and y are real numbers, and i is the imaginary number which is known as iota, whose value is $\sqrt{-1}$ .

Assume that the center of direction is,

$x_c=\frac{{\displaystyle \underset{}{\sum x_n{m}_n}}}{\sum m_n}$ ……. (1)

Substitute in equation (1), and we get,

$x_c=\frac{x_1,m_1+x_2{m}_2+x_{3}m3}{m_1+m_2+m_3}$ ........(2)

Use the 4^th assumption as:

$$\begin{gathered} x_c=\frac{m\left(x_1+x_2+x_3\right)}{3m} \\ =\frac{x_1+x_2+x_3}{3} \end{gathered}$$

Same for y :

$y_c=\frac{{\displaystyle \underset{}{\sum y_n{m}_n}}}{\sum m_n}$ ……. (3)

Substitute in equation (1), we get,

$y_c=\frac{y_1{m}_1+y_2{m}_2+y_3{m}_3}{m_1+m_2+m_3}$ ……. (4)

Use the 4^th assumption as:

$$\begin{gathered} y_c=\frac{m\left(y_1+y_2+y_3\right)}{3m} \\ =\frac{y_1+y_2+y_3}{3} \end{gathered}$$

The location equation can be written as,

$$\begin{gathered} z_c=x_c+i{y}_c \\ =\frac{x_1+x_2+x_3}{3}+i\left(\frac{y_1+y_2+y_3}{3}\right) \\ =\frac{\left(x_1+i{y}_1\right)+\left(x_2+i{y}_2\right)+\left(x_3+i{y}_3\right)}{3} \\ =\frac{z_1+z_2+z_3}{3} \end{gathered}$$

Hence the center of mass of three identical particles situated at the point$z_1,z_2,z_3$is

role="math" localid="1658743060114" $\frac{z_1+z_2+z_3}{3}$