Hello, I would very much appreciate your looking through my answer to the following question and letting me know any hick ups it may have..
Question: The points A, B, C, in an Argand diagram represent the complex numbers a, b, c, and . Prove that if λ is real then A lies on BC and divides BC in the ratio , but if is complex then, in triangle ABC, AB:BC = : 1 and angle ABC = arg .
My attempt to answer: =x+jy so a = [(1-x)+j(-y)]b+(x+jy)c. If is real then y=0 and a is simply the convex combination of b and c and lies on BC dividing it in the ratio λ: 1-λ ( shouldn't λ lie in the interval [0,1] otherwise a will lie on the extension of BC rather than on BC???) If y different to 0 then (1-) and are complex multiplying b and c respectively. This means that b - resp. c - is turned through the angle arg(1-) - resp. arg - and stretched by the scale factor "modulus of 1-" - resp. :modulus of " - to give the vector (1-)b - resp. c . Then so long as b and c are distinct ABC is a triangle and AB/AC = (a-b)/(c-b)=[(1-)b+c]/(c-b)= and therefore also the angle ABC = arg . my answer to the case where is complex sounds too confused to be right??? thanks for any help..
Made the question readable by converting to latex - Insp
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geometric interpretation of complex numbers watch
- Thread Starter
Last edited by insparato; 03-08-2009 at 22:24.
- 03-08-2009 18:17
- Thread Starter
- 03-08-2009 18:18
guys sorry about the ...
... it's meant to read "parenthesis one minus lamda close parenthesis"
- 03-08-2009 18:29
This is confusing as hell. Use latex!
- 03-08-2009 22:20
I've only had a quick look at this so bare with me.
For the first part, i would rearrange a, to something else and it should be not that hard to see why a is on BC.
For the next part, find AB and AC in terms of b and c.
letting lambda = x + iy and going through the same motions finding AB and BC in terms of b and c, you'll soon find the ratio falls out.
Haven't managed to prove <ABC = arg lambda yet, but as i said i've only taken a quick look. Hope that helps a little bit atleast.