Complex conjugate proof, i think

  • Thread starter nmsurobert
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Homework Statement



prove that sqrt2|z| greater than or equal to |Rez| + |Imz|

Homework Equations


|z|^2 = x^2 + y^2
Rez=x, Imz=y


The Attempt at a Solution


so far I've worked it down to this.
2(x^2 + y^2) greater than or equal to x^2 + 2xy + y^2

I've used a few different values for x and y and it holds true but i don't know if that right or that's how it should be written.
any input would be appreciated.
 
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Answers and Replies

  • #2
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2(x + y)^2 greater than or equal to x^2 + 2xy + y^2
Where does that come from? It is a trivial statement right now, but I think there is a mistake. Fix it, and prove the relationship with the usual rules for manipulating inequalities.
 
  • #3
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Where does that come from? It is a trivial statement right now, but I think there is a mistake. Fix it, and prove the relationship with the usual rules for manipulating inequalities.
i actually just fixed it.
 
  • #4
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Good. You can write it as inequality and reduce it to something you know to be true.
 
  • #5
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thats where im stuck. i keep trying to break it down but i keep ending up with a jumbled mess

x greater than or equal to x/2 +y - y^2/2x ?
 
  • #6
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ok i dont know if i can do this but here it is

x^2-2xy +y^2 graeter than or equal to 0

(x-y)^2 = 0

sqrt each side, move some stuff and...

x greater than or equal to y.

and thats true right?
 
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  • #7
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so...
(x-y)(x-y) greater than or equal to 0
divide one of them out
x-y greater than or equal to zero
x greater than or equal to y

i dont understand how that is intuitively true though.
 
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  • #8
Dick
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so...
(x-y)(x-y) greater than or equal to 0
divide one of them out
x-y greater than or equal to zero
x greater than or equal to y

i dont understand how that is intuitively true though.
You can't divide one of the them out without knowing the sign of x-y. And I think you might be missing a step in going from sqrt(2|z|)>=|Re(z)|+|Im(z)| to x^2-2xy-y^2>=0. Pay attention to the absolute values on Re(z) and Im(z).
 
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  • #9
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really? ((x-y)(x-y))/(x-y) doesnt equal x-y ?
 
  • #10
Dick
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really? ((x-y)(x-y))/(x-y) doesnt equal x-y ?
Yes, it does but a*a>=0 doesn't imply that a>=0.
 
  • #11
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well then... i have no idea what to do with this problem anymore.

first day of class..... completely lost.
 
  • #12
Dick
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well then... i have no idea what to do with this problem anymore.

first day of class..... completely lost.
(-1)(-1)>=0. But (-1)>=0 is false. If you divide both sides by a negative you have to reverse the inequality. You got to the correct point of wanting to prove that x^2-2xy-y^2>=0 (though you may have skipped a step, can you spell them all out). You can factor that into (x-y)^2>=0 which shows what you want to show. You don't have to (and can't!) show x>=y.
 
  • #13
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so what youre saying is (x-y)^2 >= 0 is always true. which i can see. because (x-y)^2 is always positive. ahhhh thank you.

problem 1 of 10 down haha

the last time i did proofs was in precalc 2 and those were trig proof... about 3 years ago and 5 math classes ago haha so i feel new to this stuff. its actually for a class titled "intermediate mathematical methods", PHYS395. i just havent seen this stuff in a good while.

thank you again.
 
  • #14
Dick
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so what youre saying is (x-y)^2 >= 0 is always true. which i can see. because (x-y)^2 is always positive. ahhhh thank you.

problem 1 of 10 down haha

the last time i did proofs was in precalc 2 and those were trig proof... about 3 years ago and 5 math classes ago haha so i feel new to this stuff. its actually for a class titled "intermediate mathematical methods", PHYS395. i just havent seen this stuff in a good while.

thank you again.
Ok, but what I've been trying to say is that your original statement after squaring leads to 2(x^2+y^2)>=x^2+2|x||y|+y^2. Do you see why? There's an extra bit to turn that into x^2-2xy-y^2>=0.
 
  • #15
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yeah i see why. on my paper i have it written out step by step, including that one.

i guess i was a bit unclear on when to stop the algebra haha but i see now where we stopped is the logical place to stop.
 
  • #16
Dick
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yeah i see why. on my paper i have it written out step by step, including that one.

i guess i was a bit unclear on when to stop the algebra haha but i see now where we stopped is the logical place to stop.
Ok, I'll trust you.
 
  • #17
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i can take a picture of my notes if you like. its some printing paper covered in mess though haha but its in there.

next time ill just upload a copy of my work to photobucket and post it into the "attempt at a solution" area... im sure you guys will see more of me.
my first year hitting 300 level physics courses and a 400 level math course.
thank you again.
 
  • #18
Dick
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i can take a picture of my notes if you like. its some printing paper covered in mess though haha but its in there.

next time ill just upload a copy of my work to photobucket and post it into the "attempt at a solution" area... im sure you guys will see more of me.
my first year hitting 300 level physics courses and a 400 level math course.
thank you again.
Yes, post all of the details of your solution. Always best. See you later!
 

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