# Complex conjugate proof, i think

1. Jan 17, 2015

### nmsurobert

1. The problem statement, all variables and given/known data

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

2. Relevant equations
|z|^2 = x^2 + y^2
Rez=x, Imz=y

3. 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.

Last edited: Jan 17, 2015
2. Jan 17, 2015

### Staff: Mentor

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. Jan 17, 2015

### nmsurobert

i actually just fixed it.

4. Jan 17, 2015

### Staff: Mentor

Good. You can write it as inequality and reduce it to something you know to be true.

5. Jan 17, 2015

### nmsurobert

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. Jan 17, 2015

### nmsurobert

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?

Last edited: Jan 17, 2015
7. Jan 17, 2015

### nmsurobert

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.

Last edited: Jan 17, 2015
8. Jan 17, 2015

### Dick

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).

Last edited: Jan 17, 2015
9. Jan 17, 2015

### nmsurobert

really? ((x-y)(x-y))/(x-y) doesnt equal x-y ?

10. Jan 17, 2015

### Dick

Yes, it does but a*a>=0 doesn't imply that a>=0.

11. Jan 17, 2015

### nmsurobert

well then... i have no idea what to do with this problem anymore.

first day of class..... completely lost.

12. Jan 17, 2015

### Dick

(-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. Jan 17, 2015

### nmsurobert

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. Jan 17, 2015

### Dick

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. Jan 17, 2015

### nmsurobert

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. Jan 17, 2015

### Dick

Ok, I'll trust you.

17. Jan 17, 2015

### nmsurobert

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. Jan 17, 2015

### Dick

Yes, post all of the details of your solution. Always best. See you later!

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