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Tricky complex numbers proof 
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#1
Nov805, 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. 


#2
Nov805, 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 


#3
Nov905, 07:36 AM

Sci Advisor
HW Helper
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) 


#4
Nov1705, 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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