CV19

From Example Problems
Jump to: navigation, search

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.


Main Page:Complex Variables