# geometric interpretation of complex numbers

• Aug 3rd 2009, 09:56 AM
pb6883
geometric interpretation of complex numbers
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-λ)b + λc. Prove that if λ is real then A lies on BC and divides BC in the ratio λ: 1-λ, but if λ is complex then, in triangle ABC, AB:BC = (modulus of λ) : 1 and angle ABC = argλ.

My attempt to answer: λ=x+jy so a = [(1-x)+j(-y)]b+(x+jy)c. If λ 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-λ) and λ are complex multiplying b and c respectively. This means that b - resp. c - is turned through the angle arg(1-λ) - resp. argλ - and stretched by the scale factor "modulus of 1-λ" - resp. :modulus of λ" - to give the vector (1-λ)b - resp. λc . Then so long as b and c are distinct ABC is a triangle and AB/AC = (a-b)/(c-b)=[(1-λ)b+λc]/(c-b)=λ and therefore also the angle ABC = argλ. :confused: my answer to the case where λ is complex sounds too confused to be right??? thanks for any help..
• Aug 3rd 2009, 12:49 PM
Opalg
Quote:

Originally Posted by pb6883
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-λ)b + λc. Prove that if λ is real then A lies on BC and divides BC in the ratio λ: 1-λ, but if λ is complex then, in triangle ABC, AB:BC = (modulus of λ) : 1 and angle ABC = argλ.

My attempt to answer: λ=x+jy so a = [(1-x)+j(-y)]b+(x+jy)c. If λ 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-λ) and λ are complex multiplying b and c respectively. This means that b - resp. c - is turned through the angle arg(1-λ) - resp. argλ - and stretched by the scale factor "modulus of 1-λ" - resp. :modulus of λ" - to give the vector (1-λ)b - resp. λc . Then so long as b and c are distinct ABC is a triangle and AB/AC = (a-b)/(c-b)=[(1-λ)b+λc]/(c-b)=λ and therefore also the angle ABC = argλ. :confused: my answer to the case where λ is complex sounds too confused to be right??? thanks for any help..

I think the only way you could simplify your answer a bit would be to write a = (1–λ)b + λc as a = b + λ(c–b). You can see from that formulation that multiplication by λ stretches the length of BC by a factor of |λ|, and rotates BC about B through an angle arg(λ), in order to get from B to A.
• Aug 3rd 2009, 08:54 PM
pb6883
yeah... thanks indeed, that is much better!
• Aug 4th 2009, 12:02 AM
pb6883
i hope you can find some time to help with the follwoing question also...

question: if the points a and b are two vertices of an equilateral triangle, prove that the third vertex is either

(1) b + w (b-a) or (2) b + w^2 (b-a) where w is cos(2pi/3) + j sin(2pi/3)

what i have done is to equate (1) with c and (2) with c' and then
divide th two sides of (1) and (2) and arrive at (c-b)/(c'-b) = 1/w, which tells us that vector BC' is vector BC roted about B through the angle 2pi/3 wich is true.

However I have been (Headbang) to understand why c - b = w (b-a) and why
(c'-b) = w^2 (b-a) which seems to me (having drawn a diagram) to be respectively angle pi/3 too far and angle pi/3 too short?

This is probabably obvious, so sorry/thanks in advance