Complex Numbers (1 Viewer)

[Aerath]

Just a couple of questions, if anyone would be so kind to help:

1. Prove that the three points: z₁, z₂ and z₃ are collinear, if and only if the ration (z₃-z₁) / (z₂-z₁) is real. Hence show the points 5 + 8i, 13+20i and 19 + 29i.

I think I know how to do the second bit, just gradient, right? But, not sure about the first bit

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

Purely imaginary means that for equation x +iy, x = 0 and the argument is 90 or -90 (I think). Not sure how to prove it though.

Thanks in advanced.

 

[lyounamu]

Just a couple of questions, if anyone would be so kind to help:

1. Prove that the three points: z₁, z₂ and z₃ are collinear, if and only if the ration (z₃-z₁) / (z₂-z₁) is real. Hence show the points 5 + 8i, 13+20i and 19 + 29i.

I think I know how to do the second bit, just gradient, right? But, not sure about the first bit

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

Purely imaginary means that for equation x +iy, x = 0 and the argument is 90 or -90 (I think). Not sure how to prove it though.

Thanks in advanced.

omg, how far do you accelerate on your own?

That reminds me of Foram who is doing 4 Unit at the moment. (or eliseliselise)

 

[Aerath]

Not that far ahead? We start 4U in about 2/3 months. And it's not like I'm studying by myself, anyway.

 

[Iruka]

You could solve these questions algebraically, but I think a geometric proof would probably be more enlightening.

We should think of z and w as 2D vectors lying in the Argand plane.

If (z+w) / (z-w) is pure imaginary, then arg(z+w) – arg(z-w) = +/- pi/2.

Let |z| = |w| = r. (I think we need z and w to be distinct, as well, otherwise the desired ratio is undefined.)

This means that z lies somewhere on the circle x^2+y^2 = r^2. And w is any other vector with length r. Then the three points z+w, z and z-w are collinear. (Try drawing a picture.)

Now, we can draw a second circle of radius r, centered at z, which passes through z+w, z-w and the origin. Then z+w and z-w form a diameter of this second circle, and from a theorem in circle geometry, we can see that arg(z+w) – arg(z-w) = +/- pi/2.



I think you probably should try doing the first one geometrically, too.

 

[qqmore]

for 1)i)
Since (z3 - z1) / (z2 - z1) is purely real,
therefore, arg [(z3 - z1) / (z2 - z1)] = pi or 0
=> arg (z3 - z1) - arg (z2 - z1) = pi or 0 (a)

Draw an agrand diagram with points z1,z2,z3
Join up z1 and z2 => represents z2 - z1
Join up z1 and z3 => represents z3 - z1

For the diagram to satisfy (a), clearly the gradient of z2 - z1 = gradient of z3 - z1 for the angles to subtract to 0 or pi.

Since the gradient are equal, then arg (z3 - z1) - arg (z2 - z1) = pi or 0.
Therefore, (z3 - z1) / (z2 - z1) is purely real

for ii) just let z1, z2 and z3 be 5 + 8i, 13+20i and 19 + 29i respectively and show that they end up being purely real in (z3 - z1) / (z2 - z1). Thus, by i) those points are collinear.



for 2)
(underlined letters means conjugate)
First prove that if a complex number k is purely imaginary then, k= -k.

Proof:
Let k = yi,
=> k= yi
= -yi
=-k
Therefore, k is purely imaginary if k =-k (I)

Since |z| = |w|, => |z|^2 = |w|^2 => zz = ww => z = (ww)/z (1)

(z+w) / (z-w) = (z+w)/(z-w)
= [(ww/z) +w] / [ (ww/z) -w] , by (1)
= [ww + wz] / [ww - wz]
= [w(z+w)] / [-w (z-w) ]
= - [ (z+w) / (z-w) ]
Therefore, (z+w) / (z-w)= - [ (z+w) / (z-w) ]
and by (I), clearly [(z+w) / (z-w)] is purely imaginary.

 

[Slidey]

Just a note: you don't need to prove (in fact you cant) that the conjugate of ki is -ki; it's a definition.

 

[qqmore]

But you still need to prove that if (conjugate of k) = -k, then k is purely imaginary - to use it right?