Complex Numbers Question (1 Viewer)

[Will]

Hi, I was wondering if someone could do this for me

if |Z1 + Z2| = |Z1| + |Z2|, show algebraically, that arg(Z1) = arg(Z2)


What I've done so far is

if arg(Z1) = arg(Z2)
then, arg(Z1) - arg(Z2) = 0
then, arg(Z1/Z2) = 0

ie. must show that Z1/Z2 ( R

I think this can be done by showing

Z2 - Z1 = mod(Z1)
or
Z1 + ^Z2 = mod(Z1)
or
Z2 + ^Z1 = mod(Z1)
where ^ denotes the conjugate


can anyone prove this?

 

[spice girl]

Willy, use cos rule:

|z1 + z2|^2 = |z1|^2 + |z2|^2 - 2|z1||z2|cosx

x = pi - (arg(Z2) - arg(Z1))
show this using diagram

since |z1 + z2| = |z1| + |z2|
|z1 + z2|^2 = |z1|^2 + |z2|^2 + 2|z1||z2|
cosx = -1
x = pi

arg(Z2) - arg(Z1) = 0

 

[Will]

cool,
thanks

...who are you?

 

[BlackJack]

You could say she's the smartest guy in maths in this forum...

 

[KilledInAction]

Hey I understand most of what you did, but I'm having a slight difficulty in drawing the step:

x = pi - (arg(Z2) - arg(Z1))
show this using diagram

I've made a triangle with side lengths |Z1|,|Z2| and |Z1+Z2| and I know which angle x is, but what's the best and simplest way to show x= pi - (arg(Z2) - arg(Z1) ? It doesn't seem to difficult but I'm going around in circles trying to show it

The hardest part of complex numbers for me (and most ppl I've spoken to) is definately using vectors

 

[Will]

let A be point Z1
B be point Z2 and C be point Z2 + Z1

x is the angle OAC

angle AOB = argZ1 - argZ2 (by observation)

and OAC = pi - AOB (AC||OB, cointerior angles of parallel lines supplimentary)

thus x = pi - [argZ1 - argZ2]

 

[Will]

A friend of mine also proved this result by showing Z1/Z2 is equal to its conjugate, the algebra being quite complex (heh, no pun intended).

It'd be interesting to see how many methods people come up with

 

[spice girl]

I thought it would've been obvious that if you have length of one side of a triangle equal to the sum of the lengths of other two sides, you have a degenerate triangle, with the two vectors pointing the same way.

 

[Will]

Well yes, the question asks to prove the result geometrically as well as algebraically.

obvious as the statement is,
it was the algebraic method that was beyond me.

 

[Will]

IS THAT YOU, DENG?

 

[KilledInAction]

Ahh co-interior angles of course!
Thanx for clearing that out...such a simple explanation always kills me when I've tried unnecessary working.

 

[Will]

Spicegirl:
technically, your proof is geometric as you use geometric properties (angles, parallel lines, cointerior angles of parallel lines) to give geometric representations of your algebraic constructs.

try again

 

[Will]

Here we go, algebraic proof #2 (proof #4)

let z1 = a+bi and z2=c+di
square both sides
collect like terms
square both sides again
you end up with the ratio ac=bd

ie b/a = d/c

since argZ1 = arctan (b/a) and argZ2 = arctan (d/c)

therefore argZ1 = argZ2 (as b/a = d/c)

elegant.

i will post the proof using conjugate definitions as soon as i work it out

 

[spice girl]

Elegant? How ironic! Elegance is an easy way to solve a hard problem, not the other way around.

 

[spice girl]

O and you're right btw about the algebraic method thing. I just find it stupid proving the extreme case of the triangle inequality (which is btw, another method: "By the triangle inequality, |z1 + z2| <= |z1| + |z2| {equality when z1, z2 point in the same blody direction}")

 

[spice girl]

Originally posted by Will

since argZ1 = arctan (b/a) and argZ2 = arctan (d/c)

therefore argZ1 = argZ2 (as b/a = d/c)

This bit is a little iffy...

arctan gives values between -pi/2 <= arctan(x) <= pi/2
But arg(z) gives values between -pi <= arg(z) <= pi

Clearly you've missed the case arg(z1) = arg(z2) +-pi

So I'll fix by starting that b/a = d/c
so either
1)arg(z1) = arg(z2) +-pi
or 2) arg(z1) = arg(z2)
In any case z1 = k*z2 (k is strictly real)
|z1| + |z2| = |z1 + z2|
|k|*|z2| + |z2| = |k*z2 + z2| = |(k+1)|*|z2|
(|k| + 1)|z2| = |(k+1)|*|z2|
(|k| + 1) = |(k+1)|
thus k >= 0

therefore case (2) is true.