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Vector Question (1 Viewer)

Trev

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thanx for that...
can't b bthered to make a new thread, and it doesnt relate to this thread but i cant do this complex q. - i hate vectors =|
Use the vector representation of z1 and z2 on an Argand diagram to show that:
If |z1| = |z2|, then (z1 + z2)/(z1 - z2) is imaginary.
(i can draw it and everything how they end up bisecting at 90*, i just dont get the division part! =S )
 

mojako

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LOL..
its really easy to make a new thread

since they "bisect" at 90 degrees (actually they dont bisect since z1-z2 has its tail at the origin), arg (z1 + z2)/(z1 - z2) = 90 (or -90)
because arg x/y = arg x - arg y

(z1 + z2)/(z1 - z2) = r cis(+- 90)
cos (+-90) = 0 so the real part disappears
 
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