triangle inequalities (1 Viewer)

[mr EaZy]

if we are given a qestion on complx numbers:

and argand diagram is given in a shape of a triangle (as in 95 hsc paper)
we have Z, W and |Z-W|

can we just assume that the triangle inequalites (that two added sides will always be longer than the third side) can be used to prove that |Z| +|W| > |ZW| where inequality exists only if Arg (Z) = Arg (W) = 90?

how do we answer it: just quote the triangle inequality right?
wat about part two of Q2 95 part 3?
(its the the continuation of the one i just talked about above)

thnx all!

 

[ngai]

|Z| +|W| > |ZW|

when Z = W = 3
LHS = 3+3 = 6 > RHS = 9
hence 6 > 9
yay

 

[mojako]

when Z = W = 3
LHS = 3+3 = 6 > RHS = 9
hence 6 > 9
yay

yes yes...
I'm so happy..
maths is indeed interesting
now I can start arguing with my teacher

 

[mr EaZy]

when Z = W = 3
LHS = 3+3 = 6 > RHS = 9
hence 6 > 9
yay

what, so u can prove by substituting values?
and is 6 really greater than 9 ?

What about the second part of the question i was talkin about (from the 95 paper)

where did u get 9 anyway? (im full of questions )

 

[mojako]

and is 6 really greater than 9 ?

from ur statement to be proven, yes.
thats why ur question is wrong

 

[CrashOveride]

Dude u wanna show |z| + |w| >= |z-w| not |wz|

 

[mojako]

Dude u wanna show |z| + |w| >= |z-w| not |wz|

no way!
I want 6 to be greater than 9!!
It's a breakthough in maths history!!!!

 

[withoutaface]

Dude u wanna show |z| + |w| >= |z-w| not |wz|

I think it was more:

|Z|+|W|>=|Z+W|

 

[ngai]

where did u get 9 anyway?

u gotta be kidding...
i said when W = Z = 3
then i sed RHS = 9
ur RHS is |WZ|
well, heres how i got it:
when W = 3, Z = 3
WZ = W*Z = 3*3 = 9
and since WZ = 9, |WZ| = |9| (absolute value both sides)
now, 9 is a real number, and 9 is non-negative, so |9| = 9 (since |x| = x for all non-negative and real x)
so |WZ| = |9|, and 9 = |9|
so |WZ| = 9 (both sides equal to |9|)
so RHS = 9 (since RHS = |WZ|)

 

[Jase]

im not sure where you came up with the |ZW| or arg stuff.
In the context of that question, you need to show that |z - w| <= |z| + |w|

so yes you can assume the identity : length of two sides add up to be greater than third side. so OP + OQ > PQ. also, if POQ is a straight line, then OP + OQ = PQ
hence proven for greater and equal.

ii) you just draw it in and state the obvious.

iii) if |z - w| = |z + w| then OR = PQ so the diagonals are equal. so its a rectangle.
then w = zi (OP is perpendicular to OQ). hence w/z = i or w/z is purely imaginary.

 

[mr EaZy]

Thnx Jase.

Yeah i made a mistake ots like what i said at the very top:
|Z| + |W| >= |Z-W|

But anywayz, i get it now

wait, if |Z-W| = |Z+W| , does this mean that the magnitudes of Z and W are equal? is this something we can assume with complx numbers? (since the vectors have undefined magnitudes)

 

[Xayma]

if |Z-W|=|Z+W|

Then Z=W=0 or Z=0 or W=0.

Take Z=1, W=1, their magnitude is the same
|2|=|0|
2&ne;0
&there4; false

 

[Logix]

isnt the solution to |Z-W| = |Z+W| Z=iW?

 

[Xayma]

Probably its been awhile since Ive done complex.

 

[ngai]

isnt the solution to |Z-W| = |Z+W| Z=iW?

hmm, lets see
Z = 1, W = 2i
|Z-W| = |1-2i| = sqrt5, |Z+W| = |1+2i| = sqrt5
guess not

 

[CrashOveride]

I think it was more:

|Z|+|W|>=|Z+W|

Nah it wasn't =p

if |Z-W|=|Z+W|
it just means the diagonals are equal, so yeah they are equal in magnitude and u have a rectangle.

 

[withoutaface]

Nah it wasn't =p

if |Z-W|=|Z+W|
it just means the diagonals are equal, so yeah they are equal in magnitude and u have a rectangle.

I thought he was just referring to the triangle inequality, and that is |z|+|w|>=|z+w|

 

[CrashOveride]

But you can have either =p

 

[Constip8edSkunk]

isnt the solution to |Z-W| = |Z+W| Z=iW?

its when the difference of the principle arguments of V and W is equal to pi/2