Complex Number Q (1 Viewer)

shaon0

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On an Argand diagram the points P and Q represent the complex number z1 and z2 respectively. OPQ is an equilateral triangle.
Show that (z1)^2 + (z2)^2 = 2z1.z2
Got this:
(z1-z2)^2=0
If z1=z2. then LHS=RHS.
Vector angles are : pi/3.
 
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Michaelmoo

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shaon0 said:
On an Argand diagram the points P and Q represent the complex number z1 and z2 respectively. OPQ is an equilateral triangle.
Show that (z1)^2 + (z2)^2 = 2z1.z2
Got this:
(z1-z2)^2=0
If z1=z2. then LHS=RHS.
Vector angles are : pi/3.
Yep. Looks like its right. Just make sure you put in (equilateral triangle) afte z1 = z2
 

shaon0

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Another Q:
Show that mod(mod(z1)+mod(z2)) =< mod(z1+z2)
 

kaz1

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shaon0 said:
Another Q:
Show that mod(mod(z1)+mod(z2)) =< mod(z1+z2)
You wrote the question wrong. It's ||z1| -|z2|| =< |z1 + z2| and you have to state the condition for that inequality to hold.
 

shaon0

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kaz1 said:
You wrote the question wrong. It's ||z1| -|z2|| =< |z1 + z2| and you have to state the condition for that inequality to hold.
Yeah you're right. But that doesn't solve much. Still have to prove it works and under what conditions.This is the 1st 4unit q i'm having trouble with :(
 

gurmies

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|w1| + |w2| >= |w1 + w2| [triangle inequality]

Let w1 = z1 + z2

w2 = -z2

|z1 + z2| + |-z2| >= |z1 + z2 - z2|

|z1 + z1| + |z2| >= |z1|

|z1| - |z2| <= |z1 + z2|

If we swap z1 and z2 around we get

|z2| - |z1| < = |z1 + z2|

Thus:

||z1| - |z2|| <= |z1 + z2| ...
 

shaon0

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gurmies said:
|w1| + |w2| >= |w1 + w2| [triangle inequality]

Let w1 = z1 + z2

w2 = -z2

|z1 + z2| + |-z2| >= |z1 + z2 - z2|

|z1 + z1| + |z2| >= |z1|

|z1| - |z2| <= |z1 + z2|

If we swap z1 and z2 around we get

|z2| - |z1| < = |z1 + z2|

Thus:

||z1| - |z2|| <= |z1 + z2| ...
Yeah got it before. but thanks for your help anyways.
 

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