Prove that the area A ( a, b, c ) of the triangle in the complex plane with corners at a, b, c?

Prove that the area A ( a, b, c ) of the triangle in the complex plane with corners at a, b, c


must be C (Complex), ordered in anti-clockwise fashion, is given by the formula








A ( a, b, c ) = (i/4)( ab` - a`b + b c `- b`c + c a` - c`a ) .


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(for the letters a` , b`,c` , it is actually a line on top of these letters)

Prove that the area A ( a, b, c ) of the triangle in the complex plane with corners at a, b, c?
a(a1+ia2), b(b1+ib2), a' denoting the conjugate of a,


then:


ab' = a1b1+a2b2 + i(a2b1-a1b2)


a'b = a1b1+a2b2 + i(a1b2-a2b1)


ab'-ab' = 2i(a2b1-a1b2).





Now from plane geometry you have the formula:


for area of A(a1,a2) B(b1,b2) c(c1,c2)


(ABC ordered clockwise)





Area = ½ (a1b2-a2b1) + ½(b1c2-b2c1) + ½(c1a2-c2a1)





This gives the key I hope