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Triangle inequality for complex numbers: sketch of proof

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if [itex]z_1,z_2 \in \mathbb{C}[/itex] then [itex] |z_1+z_2| \leq |z_1| + |z_2|[/itex]

    2. Relevant equations


    3. The attempt at a solution

    I tried by explicit calculation, with obvious notation for [itex]a,b[/itex] and [itex]c[/itex]: my frist claim is not that the triangle inequality holds, just that I don't know to put a ? above the [itex] \leq [/itex] symbol

    [itex] \sqrt{a} \leq \sqrt{b} + \sqrt{c} \rightarrow 0 \leq \sqrt{b}
    + \sqrt{c} - \sqrt{a} \rightarrow 0 \leq \frac{b+c+2\sqrt{b}\sqrt{c} - a}{\sqrt{b}+\sqrt{c}+\sqrt{a}}[/itex]
    Now if [itex]z_1 = x_1 + i y_1[/itex] and [itex]z_2=x_2+iy_2[/itex]

    and using again the conjugate of the roots expresion, the last equation is something like

    [itex] 0\leq f(x_1^2,x_2^2,y_1^2,y_2^2)[/itex]

    and so is true. can this be correct or may I write explicitly all the terms?

  2. jcsd
  3. Feb 21, 2012 #2


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    I really don't think that's going anywhere. You should use complex conjugates to express the absolute values. [itex]|z|^2=z \bar{z}[/itex]. Start from [itex]|z_1+z_2|^2=(z_1+z_2) (\bar{z_1} + \bar{z_2})[/itex]
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