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Note $\alpha$ and $\beta$ are complex numbers and $\overline{\alpha}$ is the conjugate of $\alpha$. I've tried using variations of the triangle inequality and I couldn't find anything to work.

  • Square them. It's much easier to work with the square of the modulus when possible. – Daniel Fischer Nov 19 '14 at 17:00
  • I forgot to mention that I have tried doing that too but I didn't succeed. –  Nov 19 '14 at 17:01
  • Okay, more hint: $\lvert 1+\overline{\alpha}\beta\rvert^2 - \lvert \alpha + \beta\rvert^2$. – Daniel Fischer Nov 19 '14 at 17:06
  • I expanded this in Maple and it gave a tremendous expression. It would be a headache to try and prove this statement with this expression. There must be a simpler way. –  Nov 19 '14 at 17:24
  • It produces in fact a relatively simple expression, provided you leave it in complex form, and don't break it up into expressions containing only the real and imaginary parts of $\alpha$ and $\beta$ (although that isn't too bad either). I'd be curious what Maple makes of it, though. – Daniel Fischer Nov 19 '14 at 17:31
  • $$\begin{aligned}\lvert 1+\overline{\alpha}\beta\rvert^2 - \lvert \alpha + \beta\rvert^2 &= 1 + \overline{\alpha}\beta + \alpha \overline{\beta} + \lvert \alpha\rvert^2\lvert\beta\rvert^2 - \lvert \alpha\rvert^2 - \overline{\alpha}\beta - \alpha\overline{\beta} - \lvert\beta\rvert^2\ &= 1 - \lvert\alpha\rvert^2 - \lvert\beta\rvert^2 + \lvert\alpha\rvert^2\lvert\beta\rvert^2\ &= (1-\lvert\alpha\rvert^2)(1-\lvert\beta\rvert^2)\end{aligned}$$ – Daniel Fischer Nov 19 '14 at 22:12

1 Answers1

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$|\alpha + \beta|^2 = |(x_1+x_2) + (y_1+y_2)i|^2 = (x_1+x_2)^2+(y_1+y_2)^2$

$|1+\bar{\alpha}\beta|^2 = |1+(x_1-y_1i)(x_2+y_2i)|^2 = |1+x_1x_2+y_1y_2 + (x_1y_2-x_2y_1)i|^2 = (1+x_1x_2+y_1y_2)^2 + (x_1y_2-x_2y_1)^2$. Thus we prove:

$x_1^2+x_2^2+2x_1x_2 + y_1^2+y_2^2+2y_1y_2 \leq 1+x_1^2x_2^2+y_1^2y_2^2 + 2x_1x_2 + 2y_1y_2 + 2x_1y_1x_2y_2 + x_1^2y_2^2 - 2x_1y_1x_2y_2 + x_2^2y_1^2$ or:

$x_1^2 + x_2^2 + y_1^2 + y_2^2 \leq 1 + x_1^2x_2^2 + y_1^2y_2^2 + x_1^2y_2^2 + x_2^2y_1^2$

or:

$x_1^2 + x_2^2 + y_1^2 + y_2^2 \leq 1 + (x_1^2+x_2^2)(y_1^2+y_2^2)$

or:

$(x_1^2+x_2^2 - 1)(y_1^2+y_2^2 - 1) \geq 0$

which is true since $|\alpha|, |\beta| \leq 1$.

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