CV19

From Exampleproblems

Show that the four points in the Argand plane represented by the complex numbers are the vertices of a square $2+i,4+3i,2+5i,3i\,$

From the given complex numbers,we can derive the points in the plane as $(2,1)(4,3)(2,5)(0,3)\,$

Let the points represent the four sides of a squate respectively $A,B,C,D\,$

Now writing the line equation $AB\,$

$\bar{AB}=(y-1)=(\frac{3-1}{4-2})(x-2)\,$

$\bar{AB}=(y-1)=(x-2)\,$

Simplifying further we get $\bar{AB}=x-y-1=0\,$

Now writing the line equation $BC\,$

$\bar{BC}=(y-3)=(\frac{5-3}{2-4})(x-4)\,$

$\bar{BC}=(y-3)=(4-x)\,$

$\bar{BC}=x+y-7=0\,$

Now writing the line equation $CD\,$

$\bar{CD}=(y-5)=(\frac{3-5}{0-2})(x-2)\,$

$\bar{CD}=(y-5)=(x-2)\,$

$\bar{CD}=-x+y-3=0\,$

Now writing the line equation $DA\,$

$\bar{DA}=(y-3)=(\frac{1-3}{2-0})(x-0)\,$

$\bar{DA}=(y-3)=(-1)(x)\,$

$\bar{DA}=x+y-3)=0\,$

The slope of $\bar{AB}=1\,$

The slope of $\bar{BC}=-1\,$

The slope of $\bar{CD}=1\,$

The slope of $\bar{DA}=-1\,$

From these slopes,the slopes of opposite sides are equal,which means they are parallel to each other and product of the one to the next is -1,which means they are perpendicular. The lenth of $AB=\sqrt{(2-4)^2+(1-3)^2}\,$

$AB=\sqrt{8}\,$

The lenth of $BC=\sqrt{(4-2)^2+(3-5)^2}\,$

$BC=\sqrt{8}\,$

The lenth of $CD=\sqrt{(2-0)^2+(5-3)^2}\,$

$CD=\sqrt{8}\,$

The lenth of $DA=\sqrt{(2-0)^2+(1-3)^2}\,$

$DA=\sqrt{8}\,$

From the above all the sides are equal. Hence all these conditions prove that the numbers form the vertices of a square.