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geometric interpretation of complex numbers watch

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    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  a = (1-\lambda) b + \lambda c . Prove that if λ is real then A lies on BC and divides BC in the ratio  (\lambda: 1-\lambda) , but if \lambda is complex then, in triangle ABC, AB:BC =  |\lambda| : 1 and angle ABC = arg  \lambda .

    My attempt to answer:  \lambda =x+jy so a = [(1-x)+j(-y)]b+(x+jy)c. If  \lambda 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 (:confused: 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- \lambda ) and  \lambda are complex multiplying b and c respectively. This means that b - resp. c - is turned through the angle arg(1- \lambda ) - resp. arg \lambda - and stretched by the scale factor "modulus of 1- \lambda " - resp. :modulus of  \lambda " - to give the vector (1- \lambda )b - resp.  \lambda c . Then so long as b and c are distinct ABC is a triangle and AB/AC = (a-b)/(c-b)=[(1- \lambda )b+ \lambda c]/(c-b)= \lambda and therefore also the angle ABC = arg  \lambda . :confused: my answer to the case where  \lambda 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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    guys sorry about the ...

    (1-&#955

    ... it's meant to read "parenthesis one minus lamda close parenthesis"
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    WT?

    This is confusing as hell. Use latex!
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    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.
 
 
 
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Updated: August 3, 2009

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