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Tricky complex numbers proof

by lektor
Tags: complex, numbers, proof, tricky
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lektor
#1
Nov8-05, 08:08 PM
P: 56
I recently was confronted by this monstrosity of a question in one of my mock exams.

|Z1 + Z2| ≤ |Z1| + |Z2|

I made a few attempts at it before becoming demoralized with the lack of progress..
|Z^2| was equal to Z1(conjugate)Z1
Hence equaling X^2 + Y^2

However even when expanding into x+iy form etc no avail, help is appreciated.
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1800bigk
#2
Nov8-05, 08:51 PM
P: 42
square both side and see if you notice anything and remember
if A=A then, surely A is less than or equal to A
Galileo
#3
Nov9-05, 07:36 AM
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P: 2,002
Geometrically this is just the triangle inequality. It just says that the sum of two sides of a triangle is always greater or equal than the third.

Since you know |z|^2=z*z. Why not write the left side out in this form?
ie: |z+w|^2=(z*+w*)(z+w)

BerkMath
#4
Nov17-05, 05:01 AM
P: 60
Tricky complex numbers proof

I'm sorry but I had to laugh when I read "I recently was confronted by this monstrosity of a question..." only to find the triangle inequality beneathe. lol.


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