Complex analysis

Question

If Z₁,  Z₂,  Z₃,  Z₄ be the 4 vertices of rhombus,  then prove that its diagonal are orthogonal... (Z_i are complex number  for i = 1,2,3,4 

 

Answer 1

 

Let Z_i = x_i + jy_i, i = 1, 2, 3, 4, represent 4 vertices of an rhombus in a clockwise order,

where x_i and y_i are real numbers and j = √(-1).

 

Z₁Z₂, Z₂Z₃, Z₃Z₄, and Z₄Z₁ are all of the same length. That is,

 

     (x₁ - x₂)² + (y₁ - y₂)² = (x₂ - x₃)² + (y₂ - y₃)²

                                     = (x₃ - x₄)² + (y₃ - y₄)²

                                     = (x₄ - x₁)² + (y₄ - y₁)²                (Eq. 1)

 

Two diagonal vectors, ^→Z₁Z₃ and ^→Z₂Z₄, are given as follows.

 

     ^→Z₁Z₃ = (x₁ - x₃) + j(y₁ - y₃)                                        (Eq. 2)

 

     ^→Z₂Z₄ = (x₂ - x₄) + j(y₂ - y₄)                                        (Eq. 3)

 

If we take an inner product of ^→Z₁Z₃ and ^→Z₂Z₄,

 

     ^→Z₁Z₃·^→Z₂Z₄ = (x₁ - x₃)(x₂ - x₄) + (y₁ - y₃)(y₂ - y₄)

 

                        = (x₁x₂ - x₂x₃ + x₃x₄ - x₄x₁) + (y₁y₂ - y₂y₃ + y₃y₄ - y₄y₁)

              

                                                                                       (Eq. 4)

 

From Eq. 1, we can derive

 

     (x₁² - 2x₁x₂ + 2x₂x₃ - x₃²) + (y₁² - 2y₁y₂ + 2y₂y₃ - y₃²) = 0

 

                                                                                       (Eq. 5)

 

     (x₃² - 2x₃x₄ + 2x₄x₁ - x₁²) + (y₃² - 2y₃y₄ + 2y₄y₁ - y₁²) = 0

 

                                                                                       (Eq. 6)

 

If we add Eq. 5 and Eq. 6, and divide it by -2, then we get

 

    

     (x₁x₂ - x₂x₃ + x₃x₄ - x₄x₁) + (y₁y₂ - y₂y₃ + y₃y₄ - y₄y₁) = 0

 

 

As ^→Z₁Z₃·^→Z₂Z₄ = 0, two diagonals of a rhombus is orthogonal to each other.