Question

Show that the points in the Argand plane represented by the complex numbers
 \[ - 2 + 7i, - \dfrac{3}{2} + \dfrac{1}{2}i,4 - 3i,\dfrac{7}{2}(1 + i)\] are the vertices of a rhombus.

Answer 1

Hint: First of all convert the given complex numbers in the form of the points and will find the distance between the two points and will verify all the four sides of the quadrilateral are equal and same or not. As in rhombus all the four sides of the rhombus are equal.

Complete step by step solution:
Let all the four given complex numbers be represented by –
 $
  A = ( - 2,7) \\
  B = ( - \dfrac{3}{2},\dfrac{1}{2}) \\
  C = (4, - 3) \\
  D = (\dfrac{7}{2},\dfrac{7}{2}) \;
  $
Now, applying the distance formula, find measures of all the four sides –
 $ AB = \sqrt {{{( - 2 + \dfrac{3}{2})}^2} + {{(7 - \dfrac{1}{2})}^2}} $
Simplify the above expression using the concept that when there is one positive term and negative terms while simplifying you have to subtract and give sign of the bigger number.
 $ AB = \sqrt {{{( - \dfrac{1}{2})}^2} + {{(\dfrac{{13}}{2})}^2}} $
Simplify the above expression –
 $ AB = \sqrt {\dfrac{1}{4} + \dfrac{{169}}{4}} $
When bases are the same, numerators are added directly.
 $ AB = \sqrt {\dfrac{{170}}{4}} $
Apply the concept of square root –
 $ AB = \dfrac{1}{2}\sqrt {170} $ ….. (A)
Similarly,
 $ BC = \sqrt {{{( - \dfrac{3}{2} - 4)}^2} + {{(\dfrac{1}{2} + 3)}^2}} $
Simplify the above expression using the concept, when there are two negative signs you have to add both the terms and give sign of negative to the resultant value.
 $ BC = \sqrt {{{( - \dfrac{{11}}{2})}^2} + {{(\dfrac{7}{2})}^2}} $
Simplify the above expression-
 $ BC = \sqrt {\dfrac{{121}}{4} + \dfrac{{49}}{4}} $
When bases are the same, numerators are added directly.
 $ BC = \sqrt {\dfrac{{170}}{4}} $
Apply the concept of square root –
 $ BC = \dfrac{1}{2}\sqrt {170} $ ….. (B)
Similarly,
 $ CD = \sqrt {{{(4 - \dfrac{7}{2})}^2} + {{( - 3 - \dfrac{7}{2})}^2}} $
Simplify the above expression –
 $
  CD = \sqrt {{{(\dfrac{1}{2})}^2} + {{( - \dfrac{3}{2})}^2}} \\
  CD = \sqrt {\dfrac{1}{4} + \dfrac{{169}}{4}} \\
  $
When bases are the same, numerators are added directly.
 $ CD = \sqrt {\dfrac{{170}}{4}} $
Apply the concept of square root –
 $ CD = \dfrac{1}{2}\sqrt {170} $ ….. (C)
Similarly,
 $ DA = \sqrt {{{(\dfrac{7}{2} + 2)}^2} + {{(\dfrac{7}{2} - 7)}^2}} $
Simplify the above expression –
 $
  DA = \sqrt {{{(\dfrac{{11}}{2})}^2} + {{( - \dfrac{7}{2})}^2}} \\
  DA = \sqrt {\dfrac{{121}}{4} + \dfrac{{49}}{4}} \;
  $
When bases are the same, numerators are added directly.
 $ DA = \sqrt {\dfrac{{170}}{4}} $
Apply the concept of square root –
 $ DA = \dfrac{1}{2}\sqrt {170} $ ….. (D)
From equations (A), (B), (C) and (D), all the four sides are equal and same.
Hence, A, B, C, D form the Rhombus.

Note: Be careful about the sign convention while simplification. Always remember when there are two different signs then you have to do subtraction and sign of a bigger number to the resultant value and when there are two same signs do addition and given common sign of the numbers.