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Remmert chapter 1 page 13 and 14

Set $z:= x+iy, w:= u+iv$: $$\langle z,w\rangle^2+\langle iz,w\rangle^2 = Re(w\bar{z})^2+Re(iw\bar{z})^2=(ux+vy)^2+(uy-vx)^2= (x^2+y^2)(u^2+v^2)= |z|^2|w|^2$$

so from this it follows that: $\langle z,w\rangle^2 \le \langle z,w\rangle^2+\langle iz,w\rangle^2 = |z|^2|w|^2 \Rightarrow \langle z,w\rangle \le |z||w|$

If we put: $$|z+w|^2 = |z|^2+ 2\langle z,w\rangle+ |w|^2 \le |z|^2+2|zw|+|w|^2 = (|z|+|w|)^2$$

we can also say that: $|z+w| \le |z|+|w|$

lastly we can put $|z|= |z+w-w| \le |z-w|+|w| \Leftrightarrow |z|-|w| \le |z-w| \Rightarrow ||z|-|w||\le ||z-w||= |z-w|$

Have we done this correctly? Please, do tell.

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2  
Just a TeX note: It's better to use \langle and \rangle for such brackets: $\langle z, w\rangle$. The spacing is better. Also, it's hard for me to figure out what you're trying to show. –  Dylan Moreland Feb 10 '12 at 15:28
    
@Dylan Moreland, I want show this 1 equality and then 3 inequalities. –  VVV Feb 10 '12 at 15:52
1  
I changed $<z,w>$ to $\langle z, w \rangle$. While I was at it I changed all the occurrences of ^{2} to ^2. Those needless braces seem like clutter that makes it inconvenient to edit. –  Michael Hardy Feb 10 '12 at 19:20
    
Thank you, Michael Hardy. –  VVV Feb 11 '12 at 11:56

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