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Complex Numbers Question (1 Viewer)

bored of sc

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It's from the 2008 CSSA Trial.

2 (e) The points O, A, Z and C on the Argand diagram represent complex numbers 0, 1, z and z+1 respectively, where z = cos@ + isin@ is any complex number of modulus 1, with 0 < @ < pi.

(i) Explain why OACZ is a rhombus.
(ii) Show that (z-1)/(z+1) is purely imaginary.
(iii) Find the modulus and agrument of z+1.

For (i) is it correct/enough in saying A + Z = diagonal of rhombus from O (by definition) = 1+z = C or should other sides and angles be explained too (it's for 1 mark by the way).

(ii) Nop idea. They are the diagonals of the rhombus and bisect each other at 90. But how does that help with the solution.

(iii) For the modulus do you use cosine rule with sides of rhombus, and angle OAC (which can be found by using angle sum of triangle and knowing the argument). And is the argument simply @/2?


Please help. I really need 50% in this trial on Monday.
 
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GUSSSSSSSSSSSSS

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i) yeh tbh that's wat i wud do

ii) they ARE the diagonals of a rhombus and YES THEY DO interesect each other at right angles
and z-1 is one diagonal, z+1 is the other
therefore arg((z-1)/(z+1)) = 90 ((z-1)/(z+1) is the intersection of z-1 and z+1)
purely img numbers have arg = 90
therefore purely imaginary

iii) yep that seems reasonable enough to me xD
 

bored of sc

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Thanks guys. I'm freaking out cause I did the 2007 CSSA Trial Q1 and it was actually quite hard.

I need to nail most of Q's 1-4 to have a chance at 50% I reckon.

Good luck to you guys too. When are your extension 2 exams?
 

untouchablecuz

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It's from the 2008 CSSA Trial.

2 (e) The points O, A, Z and C on the Argand diagram represent complex numbers 0, 1, z and z+1 respectively, where z = cos@ + isin@ is any complex number of modulus 1, with 0 < @ < pi.

(i) Explain why OACZ is a rhombus.
(ii) Show that (z-1)/(z+1) is purely imaginary.
(iii) Find the modulus and agrument of z+1.

For (i) is it correct/enough in saying A + Z = diagonal of rhombus from O (by definition) = 1+z = C or should other sides and angles be explained too (it's for 1 mark by the way).

(ii) Nop idea. They are the diagonals of the rhombus and bisect each other at 90. But how does that help with the solution.

(iii) For the modulus do you use cosine rule with sides of rhombus, and angle OAC (which can be found by using angle sum of triangle and knowing the argument). And is the argument simply @/2?


Please help. I really need 50% in this trial on Monday.
for iii)

what you could do is:

let the point of intersection of the diagonals be D

angle DOA=@/2 (diagonals bisect the angles of a rhombus)

angle ADA=90, hence triangle ODA is right

hence,

cos(@/2)=OD/1=OD

|z+1|=OD+DC=2OD, (diagonals bisect eachother)

hence |z+1|=2cos(@/2)
 

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