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Prove |z+w|<=|z|+|w| (1 Viewer)

member 6003

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you have to start with this:
where z, w are any complex numbers
1664508312901.png
And prove:
1664508452481.png
My method:
let Z,W be conjugates of z,w
1664508312901.png
1664508621526.png - expand RHS
1664508646787.png
then if the thing is true then:
1664508902122.png
square both sides:
1664508800371.png
1664508820888.png
and then I'm stuck. I saw another guy do the question on stack exchange and they basically got to the same step but then squared both sides again (letting z=a+ib, w=x+iy) and whatever, then factorise so that its like x^2>=0 therefore must be true.
But I don't understand how you can square both sides at this step since LHS can be negative, and RHS must be positive. I thought you could only square both sides of an inequality when both sides are positive.
This question is out of extension 2 Cambridge 1D question Q25 b)
 

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idkkdi

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you have to start with this:
where z, w are any complex numbers
View attachment 36399
And prove:
View attachment 36400
My method:
let Z,W be conjugates of z,w
View attachment 36399
View attachment 36402 - expand RHS
View attachment 36403
then if the thing is true then:
View attachment 36407
square both sides:
View attachment 36405
View attachment 36406
and then I'm stuck. I saw another guy do the question on stack exchange and they basically got to the same step but then squared both sides again (letting z=a+ib, w=x+iy) and whatever, then factorise so that its like x^2>=0 therefore must be true.
But I don't understand how you can square both sides at this step since LHS can be negative, and RHS must be positive. I thought you could only square both sides of an inequality when both sides are positive.
This question is out of extension 2 Cambridge 1D question Q25 b)
Square finish the proof then write it backwards and it should be legit.
 

Trebla

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A corollary you need first is that for any complex number z (something you can easily prove):


Notice that since the terms of the sum are conjugates of each other:
 

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