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Vectors.. (1 Viewer)

Just.Snaz

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I hate them.. :(

I drew what I have so far, and I know you can prove it's a rhombus but then I don't know what to do next. I think there's a rule I'm missing..

Any help appreciated =]
 

Trebla

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arg [(z1+z2)/(z1-z2)] = arg (z1+z2) - arg(z1-z2)
From your rhombus, you know the diagonal meet at right angles, so just translate vector AB to the origin and prove that the difference of their arguments is 90 degrees.
 

Just.Snaz

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Trebla said:
arg [(z1+z2)/(z1-z2)] = arg (z1+z2) - arg(z1-z2)
From your rhombus, you know the diagonal meet at right angles, so just translate vector AB to the origin and prove that the difference of their arguments is 90 degrees.
This I don't understand. Sorry, I really suck at vectors..
 

vds700

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Just.Snaz said:
This I don't understand. Sorry, I really suck at vectors..
for a complex number to be purely imaginary, it must have an argument of pi/2 or -pi/2
 

Just.Snaz

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vds700 said:
for a complex number to be purely imaginary, it must have an argument of pi/2 or -pi/2
No sorry, I mean I don't understand the translating the vector AB bit..
 

midifile

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On the diagram draw a line parallel to the line z1 - z2, with what was z1 at the origin (so the line is going into the second quadrant - althouhg you are gonna have to expand your x axis so you have a second quadrant).

The you just have to prove that the angle between the new line you have drawn and z1 + z2 is pi/2, which is easy, as the new line youve drawn is parallel to z1 - z2
 

vds700

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Just.Snaz said:
No sorry, I mean I don't understand the translating the vector AB bit..
Complex no.s as vectors can be represented in the form rcis(theta) wher r is the length and theta is the angle they make with the positive x axis. You can shift vectors around the complex plane AS long as u dont change the direction or magnitude.
 

Just.Snaz

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midifile said:
On the diagram draw a line parallel to the line z1 - z2, with what was z1 at the origin (so the line is going into the second quadrant - althouhg you are gonna have to expand your x axis so you have a second quadrant).

The you just have to prove that the angle between the new line you have drawn and z1 + z2 is pi/2, which is easy, as the new line youve drawn is parallel to z1 - z2
ohhhh... I get it... thank you very much :D

thanks for your help everyone :eek:
 

Affinity

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I will use z and w for z1 and z2 and z' w' to denote their conjugates

(z+w)/(z-w)
= (z+w)(z'-w') / |z-w|^2
= (|z|^2 - |w|^2 - zw' + wz') / |z-w|^2
----since |z| = |w|-----
= (wz' - zw')/|z-w|^2
= [wz' - (wz')']/|z-w|^2
which must be pure imaginary because it's the difference between a number and it's conjugate
 

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