complex nos. vectors question (1 Viewer)

[uraleech]

Hi guys!

Vectors has always been the weakest part of complex numbers for me, so a little help would be appreciated in confirming the solution to this question:

The four complex numbers z1, z2, z3, z4 are represented by A, B, C and D respectively.
If z1 - z2 + z3 - z4 = 0 and z1 -iz2 -z3 + iz4 = 0
determine the possible shape(s) for quadrilateral ABCD, providing reasons.

What I've got is that opposite sides are equal, but I'm not sure how to proceed from there.

Thanks in advance.

 

[Riviet]

Hello there.

Hmm... an interesting vector question. Maybe try testing all the quadrilaterals and see which ones satisfy the two conditions.

Here are some rearranged versions of the two conditions which you may or may not have encountered and might help:

z₁-z₂ = z₄-z₃

z₁-z₄ = z₂-z₃

z₁-z₃ = i(z₂-z₄)

 

[uraleech]

Hello there.

Hmm... an interesting vector question. Maybe try testing all the quadrilaterals and see which ones satisfy the two conditions.

Here are some rearranged versions of the two conditions which you may or may not have encountered and might help:

z₁-z₂ = z₄-z₃

z₁-z₄ = z₂-z₃

z₁-z₃ = i(z₂-z₄)

aah many thanks. It was the third one which I needed. Nothing particularly special struck me until I saw it in factorised form.

 

[Riviet]

So what's your answer?

 

[Mountain.Dew]

answering uraleech's question:
a diagram would help A LOT. on a number plane, i just place one complex number on each quadrant, starting from 1st quadrant with z1, 2nd quad with z2, and so on.

we have these conditions:

z1 - z2 + z3 - z4 = 0 and z1 -iz2 -z3 + iz4 = 0

1) z1 - z2 + z3 - z4 = 0, we rearrange to get:

(z1 -z4) = (z2 -z3)
hence |(z1 -z4) | = |(z2 -z3)| and arg(z1 -z4) = arg(z2 -z3)

So, in quad ABCD, we know that it has two parallel equal sides.

similarily, from z1 - z2 + z3 - z4 = 0,
z1-z2 = z4 - z3, and also applying modulus and argument,

we now know that we have a pair of parrallel, equal sides. so its a parallelogram. NOT YET.

z1 -iz2 -z3 + iz4 = 0, rearrange to get: z1-z3 = i(z2-z4)

So, we know that the diagonals of ABCD are PERPENDICULAR.

HENCE, by properties of perpendicular diagonals and parallel, equal sides, ABCD is a RHOMBUS.

if u feel that there is any problems, reply to this thread or PM me.

 

[Riviet]

I was about to ask something, but I suddenly just understood how you deduced that the diagonals were perpendicular.

Thanks Mountain Dew.

 

[uraleech]

the answer I got was that it is either a square or a rhombus.

from z1 - z2 + z3 - z4 = 0
opposite sides are equal

from z1 -iz2 -z3 + iz4 = 0
diagonals are perpendicular

the two quadrilaterals with those properties are the square and rhombus.

 

[KeypadSDM]

the answer I got was that it is either a square or a rhombus.

from z1 - z2 + z3 - z4 = 0
opposite sides are equal

from z1 -iz2 -z3 + iz4 = 0
diagonals are perpendicular

the two quadrilaterals with those properties are the square and rhombus.

Actually it's just the rhombus. We don't include subsets in the answer to questions, it's silly.

 

[Riviet]

Keypad, do you mean like a square is a rhombus but a rhombus is not a square? Hence the rhombus is enough?

 

[YBK]

Keypad, do you mean like a square is a rhombus but a rhombus is not a square? Hence the rhombus is enough?

A square is a special rhombus i think.

all sides equal, rhombus.
all sides equal and they meet at right angles, square.

 

[Templar]

A square is a rhombus, hence the set of square is a subset of set of rhombus.

 

[uraleech]

hold on a second; i just thought of something

since z1 -iz2 -z3 + iz4 = 0
therefore (z1 - z3) = i(z2-z4)

we all agree that the diagonals are perpendicular. and from the other condition, that opposite sides are equal.

but, rotation by i does not affect the length of a vector. therefore, doesn't that mean that the diagonals are equal in length?

if diagonals are equal in length, diagonals are perpendicular, and opposite sides are equal in length, then it must be a square, since this is the only quadrilateral which has all three of these properties.

right? or did i miss something?