Help with Complex Numbers (1 Viewer)

[the-derivative]

Hi guys,

I think this is the first time I've actually posted for help with a maths problem - but I can't seem to see this question - stupid vectors (and they're probably easy to most people)

Anyway,

1. Prove that three points z1, z2 and z3 are collinear if only the ratio (z3 - z1) / (z2 - z1) is real.

2. If |z| = |w|, prove that (x+w)/(z-w) is purely imaginary.

Thanks guys.

 

[youngminii]

Okay now don't quote me on this (there's probably a heaps faster way as well)
But for 1.
Let z1 = x1 + iy1
z2 = x2 + iy2
z3 = x2 + iy3

So, (z3 - z1)/(z2 - z1)
= (x3 + iy3 - x1 - iy1)/(x2 + iy2 - x1 - iy1)
For this to be Real
(from the numerator)
iy3 - iy1 = 0
Taking out the i's
y3 - y1 = 0___(1)

(from the denominator)
iy2 - iy1 = 0
Taking out the i's
y2 - y1 = 0___(2)

(1) - (2)
y3 - y2 = 0___(3)


Okay so on the Argand Plane, Let
z1 = A
z2 = B
z3 = C

If they're collinear, mAB = mBC = mAC
m is the gradient (rise over run)
mAB = (y2 - y1)/(x2 - x1)
= 0/(x2 - x1) (from (2))
= 0
mBC = (y3 - y2)/(x3 - x2)
= 0/(x3 - x2) (from (3))
= 0
mAC = (y3 - y1)/(x3 - x1)
= 0/(x3 - x1) (from (1))
= 0
And since the gradient is all the same, they're all collinear (and in fact, on a line parallel to the x axis)

That might just be an extremely long way to do it, if so, sorry :S

 

[youngminii]

Hi guys,

I think this is the first time I've actually posted for help with a maths problem - but I can't seem to see this question - stupid vectors (and they're probably easy to most people)

Anyway,

1. Prove that three points z1, z2 and z3 are collinear if only the ratio (z3 - z1) / (z2 - z1) is real.

2. If |z| = |w|, prove that (x+w)/(z-w) is purely imaginary.

Thanks guys.

For 2. Are you sure it's x + w on the numerator? 'Cause when I draw it on the Argand Plane, it doesn't seem to work :S

 

[azureus88]

for the first one, i think you can just do this:

let z1 = OA, z2 = OB z3 = OC

z3-z1=CA and z2-z1=BA and if they have the same gradient (that is, if they are collinear) then:

arg(z3-z1) = arg(z2-z1)
so arg (z3-z1/z2-z1) = 0

therefore, z3-z1/z2-z1 is real

As for the 2nd one, is x meant to be a complex number?

 

[youngminii]

for the first one, i think you can just do this:

let z1 = OA, z2 = OB z3 = OC

z3-z1=CA and z2-z1=BA and if they have the same gradient (that is, if they are collinear) then:

arg(z3-z1) = arg(z2-z1)
so arg (z3-z1/z2-z1) = 0

therefore, z3-z1/z2-z1 is real

As for the 2nd one, is x meant to be a complex number?

For one thing, I don't think that makes sense
Second, I thought you weren't allowed to assume that the statement is true? 'Cause what you did was use the statement to prove what's given to you.
Especially in a question with 'only if'

Correct me if I'm wrong

 

[azureus88]

For one thing, I don't think that makes sense
Second, I thought you weren't allowed to assume that the statement is true? 'Cause what you did was use the statement to prove what's given to you.
Especially in a question with 'only if'

Correct me if I'm wrong

yeh, sry i misread the question. but i think it works the other way as well:

if z3-z1/z2-z1 is real, then arg(z3-z1/z2-z1) = 0 or pi

if arg(z3-z1/z2-z1) = 0, then arg(z3-z1)=arg(z2-z1) and so they are collinear as the vectors point in the same direction.
if arg(z3-z1/z2-z1) = pi, then arg(z3-z1)=pi + arg(z2-z1) and they are also collinear as the vectors lie on same line but point in opposite directions.

btw, why dont you think it makes sense?

 

[the-derivative]

Thanks heaps youngminii and azureus88!

I think I'll be using youngminii's solutions for the first question - because I also don't think you're allowed to assume that the statement is true.

Again thanks heaps guys.