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Math Help - [SOLVED] Inequality

  1. #1
    MHF Contributor arbolis's Avatar
    Apr 2008

    [SOLVED] Inequality

    If C is the frontier of the triangle with vertices z_1=0, z_2=3i and z_3=-4, show that \left| \int _C (e^z- \overline z)dz \right| \leq 60 without calculating the integral.

    My attempt : I drew the sketch of C.
    Then I wrote \left| \int _C (e^z- \overline z)dz \right| \leq \int _C  \left| (e^z- \overline z) \right| dz.
    I also wrote that \int _C dz = \sqrt{4^2+3^3}+4+3=12, hence I'm done if I show that |e^z- \overline z|\leq 5 for z \in C. (Correct me please if I'm wrong saying that z \in C, where C denotes the frontier of the triangle.)
    Here's where I'm stuck. I thought about calculating the image of the interior of the triangle under the transformation z \mapsto e^z but I believe this is a long work. Also it wouldn't finish there. I believe I'm missing something obvious.
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  2. #2
    Sep 2009
    Everything is fine with your idea, arbolis. Now, in order to prove that |e^{z}-\overline z| is less than or equal to 5 over C, all you have to do is apply the triangle inequality. You get |e^{z}-\overline z|  \leq |e^{z}|+|z|. Since |e^{z}| is equal to e^{\mathrm{Re}(z)} and e^{\mathrm{Re}(z)} \leq 1 over C, you conclude that |e^{z}-\overline z| \leq 1 + |z| \leq 1 + 4 =5. Done.
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