1. ## Complex Numbers II

Thanks for the help on my previous question, but I am struggling to find a way to solve the following problem:

Prove that the area A(a,b,c) of the triangle in the complex plane with corners at a,b,c EC, ordered in anti-clockwise fashion, is given by the formula:

A(a,b,c)= i/4*(a*b'-a'*b+b*c'-b'*c+c*a'-c'*a).

Many thanks to anyone who is able to help.

2. Hello, smiler!

Prove that the area $A(a,b,c)$ of the triangle in the complex plane
with corners at $a,b,c \in C$, ordered in anti-clockwise fashion, is given by the formula:

$A(a,b,c) \;= \;\frac{i}{4}(aB-Ab+bC-Bc+cA-Ca)$

What are $A,B,C$ ?
. . (You've already used $A$ to represent the area.)

That expression comes from: . $\frac{i}{4}\begin{vmatrix}1 & 1 & 1 \\ a & b & c \\ A & B & C \end{vmatrix}$

. . but I have no idea what I'm talking about . . .

3. ## clarification

sorry i forgot to include that:

A denotes the letter a prime (a'),
B denotes b prime (b')
C denotes c prime (c')

where aB implies a is multiplied by b' likewise for other components.

thanks

4. Do you mean complex conjugates?

5. ## reply to jane bennett

i am not totally sure what a' b' or c' mean but if you are wondering what a'*b means this means a' multiplied by b.

i am not able to ask my tutor what they mean as obviously it is the christmas holiday.

thanks

6. Hello, smiler!

I proved it . . . the looong way.

Prove that the area of the triangle in the complex plane with vertices
at $a,b,c \in C$, ordered in CCW fashion, is given by the formula:

. . $A \:= \:\frac{i}{4}\left(a b'-a' b +b c' - b' c + c a' - c' a\right)$
Theorem
The triangle with vertices: $P(x_1,y_1),\,Q(x_2,y_2),\,R(x_3,y_3)$ has area given by:

. . $\Delta PQR \;=\;\frac{1}{2}\begin{vmatrix}1 & 1 & 1 \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{vmatrix} \;=\;\frac{1}{2}\left(x_1y_2 - x_2y_1 + x_3y_1 - x_1y_3 + x_2y_3 - x_3y_2\right)\;\;{\color{blue}[1]}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Let: . $\begin{array}{ccccccc}a & = & x_1+iy_2 & & a' &=&x_1 - iy_1\\ b & = & x_2 + iy_2 & & b' &=&x_2-iy_2\\ c & = & x_3+iy_3 & & c' &=&x_3-iy_3\end{array}$

$ab' - a'b \;=\;(x_1+iy_1)(x_2-iy_2) - (x_1-iy_1)(x_2+iy_2)$
. . . . . . $= \;x_1x_2 - ix_1y_2 + ix_2y_1 + y_1y_2 - x_1x_2 - ix_1y_2 + ix_2y_1 - y_1y_2$
. . . . . . $= \;-2i(x_1y_2 - x_2y_1)$

$bc' - b'c \;=\;(x_2+iy_2)(x_3-iy_3) - (x_2-iy_2)(x_3+iy_3)$
. . . . . . $= \;x_2x_3 - ix_2y_3 + ix_3y_2 + y_2y_3 - x_2x_3 - ix_2y_3 + ix_3y_2 - y_2y_3$
. . . . . . $= \;-2i(x_2y_3 - x_3y_2)$

$ca' - c'a \;=\;(x_3+iy_3)(x_1-iy_1) - (x_3-iy_3)(x_1+iy_1)$
. . . . . . $= \;x_1x_3 - ix_3y_1 + ix_1y_3 + y_1y_3 - x_1x_3 - ix_3y_1 + ix_1y_3 -y_1y_3$
. . . . . . $= \;-2i(x_3y_1 - x_1y_3)$

$(ab' - a'b) + (bc' - b'c) + (ca' - c'a) \;=\;-2i(x_1y_2-x_2y_1) - 2i(x_2y_3-x_3y_2) - 2i(x_3y_1 - x_1y_3)$

. . . . . . . . . . . . . . . . . . . $= \;-2i(x_1y_2-x_2y_1 + x_3y_1-x_1y_3 + x_2y_3 - x_3y_2)$

Therefore:

$\frac{i}{4}(ab' - a'b + bc' - b'c + ca' - c'a) \;=\;\frac{i}{4}(-2i)(x_1y_2-x_2y_1 + x_3y_1-x_1y_3 + x_2y_3-x_3y_2)$

. . . . . . . . . . . . . . . . . $= \;\frac{1}{2}(x_1y_2-x_2y_1 + x_3y_1-x_2y_3 + x_2y_3-x_3y_2)$

which equals the formula in [1] . . . Q.E.D.