Complex vector q (1 Viewer)

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for ii is it fine if i did the following

z3-z2=eb
z3-z1=ac

Let m be the mid point of bc then ab=mc therefore eabc will form a kite
angle amb=pi/2

arg <amb=arg(z2-z5)-arg(z3-z1)= arg ((z2-z5)/(z3-z1))=pi/2 therefore ((z2-z5)/(z3-z1)) is purely imaingary

 

[Drongoski]

For (ii) let the "centre" of the regular hexagon be M ( M is also the common mid-points of AD, BE & CF). All the triangles MAB, MBC, MCD, MDE, MEF & MFA are equilateral. The figure, MABC is made up of triangles MAB & MBC is a rhombus so their diagonals AC & MB are perpendicular to each other. That means z2-z5 is perpendicular to z3-z1.

Therefore arg

Hence must be a purely imaginary number.

 

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Ohh i see so would the argument of it being a kite not work?

 

[Drongoski]

Ohh i see so would the argument of it being a kite not work?

The diagonals of a kite also intersect at 90 degrees; so should work.
MABC here is both a kite(a kite does not normally have all 4 sides equal; one pair of sides is usually shorter than the other pair) and a rhombus.

 

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ohh makes sense thanks