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Thread: Complex Analysis: Prove this Inequality

  1. #1
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    Complex Analysis: Prove this Inequality

    Prove that for $\displaystyle z \in \mathbb{C}$, $\displaystyle |z| \leq |Re \ z| + |Im \ z| \leq \sqrt{2}|z|$.
    Last edited by joestevens; Sep 6th 2010 at 08:19 AM.
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  2. #2
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    Surely you that $\displaystyle |a|\le |a|+ |b|$. You also need to know that $\displaystyle 2|a||b|\le |a|^2+ |b|^2 $.

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

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

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