Is it possible to reduce this inequality to (|x| - |y|)² ≥ 0?

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One Question...Complex Numbers

Homework Statement



verify that: Square root (2) * lzl >= l Re (z) l + l I am ( z ) l


suggestion : reduce this inequality to ( lxl - lyl )^2 >=0

note : lxl <<<< modulus x
 
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Well, what have you TRIED? If z= x+ iy, what is |z|? Whar are Re(z) and Im(z)?

I assume your note simply means that |x| and "modulus x" are the same thing. I had first interpreted it as "much less than" which makes no sense!
 


lxl <<<< modulus x<<<< i dosent mean less thab but i mean modulus x ---->lxl
i mean modulus x = lxl

((2x2+2y2)1/2 - lxl - lyl )2 >= 0

Substitute lxl - lyl = k
2x2 + 2y2 - 2k * (2x2 + 2y2)1/2 + k2 >= 0

Substitute 2x2 + 2y2 = m

m - 2k * m1/2 + k2 >= 0

this is my try ...
i know its wrong ...
and i can't continue ...
 
Last edited:


You're going the wrong way. From your attempt it looks like you know |z|=sqrt(x^2+y^2) for z=x+iy. Start with what you want to prove. sqrt(2)*sqrt(x^2+y^2)>=|x|+|y|. Square both sides. Now move everything to one side and look at your suggestion.
 
Welcome to PF!

Hi selena! Welcome to PF! :smile:

(have a square root: √ and a square: ² :wink:)

Do you mean 2(x² + y²) ≥ (|x| + |y|)²?

hmm … let's keep it simple … :rolleyes:

Hint: 2(x² + y²) = |x|² + |x|² + |y|² + |y|² :smile:
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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