step in proof: Cauchy-Scharz complex inequality

Question

Is there a smart way to see this? I tried writing it out for $n=1$ ($z=x+iy,w=u+iv$), expecting it to be simple, but I got: $$ (x-\operatorname{Re}(\lambda)u+\operatorname{Im}(\lambda)v)^2+(y-\operatorname{Im}(\lambda)u+\operatorname{Re}(\lambda)v)^2 $$ Halfway throughout the expansion, I feel like this can't be the way to go? Is there a simpler approach?

score 0 · Accepted Answer · answered Apr 11 '18 at 12:44

0

With the substitution $a = z_j$ and $b= \lambda \overline w_j$ the desired identity becomes $$ | a - b |^2 = |a|^2 + |b|^2 - 2 \operatorname{Re}(a \overline b) $$ and that follows easily by expanding $$ | a - b |^2 = (a - b)\overline{(a - b)} = a \overline a + b \overline b - a \overline b - b \overline a \, . $$

answered Apr 11 '18 at 12:44

Martin R

113,040

oh gees, I completely forgot to use $\vert z\vert^2=z\overline z$. Thanks, that does the trick! – Sha Vuklia Apr 11 '18 at 12:46
@ShaVuklia: Yes, that is sometimes easier to handle than cartesian coordinates. – Martin R Apr 11 '18 at 12:48

step in proof: Cauchy-Scharz complex inequality

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