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Math Help - Complex Analysis: Prove this Inequality

  1. #1
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    Complex Analysis: Prove this Inequality

    Prove that for z \in \mathbb{C}, |z| \leq |Re \ z| + |Im \ z| \leq \sqrt{2}|z|.
    Last edited by joestevens; September 6th 2010 at 09:19 AM.
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  2. #2
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    Surely you that |a|\le |a|+  |b|. You also need to know that  2|a||b|\le |a|^2+  |b|^2 .

    Look at \left( {\left| a \right| + \left| b \right|} \right)^2  \leqslant \left| a \right|^2  + 2\left| a \right|\left| b \right| + \left| b \right|^2  \leqslant 2\left| a \right|^2  + 2\left| b \right|^2

    Taking the square root, \left( {\left| a \right| + \left| b \right|} \right) \leqslant \sqrt 2 \sqrt {\left| a \right|^2  + \left| b \right|^2 }

    Now let a=\text{Re}(z)~\&~ b=\text{Im}(z).
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  3. #3
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    thanks.

    I made a mistake when i typed the question. i had |x| \leq ... . it should be |z| \leq .... i changed it in the original post.
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  4. #4
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    Quote Originally Posted by joestevens View Post
    thanks.

    I made a mistake when i typed the question. i had |x| \leq ... . it should be |z| \leq .... i changed it in the original post.
    That makes no difference \sqrt {\left| a \right|^2  + \left| b \right|^2 }  \leqslant \sqrt {\left| a \right|^2  + 2\left| {ab} \right| + \left| b \right|^2 }  = \left| a \right| + \left| b \right|
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