Exercise 1.13

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Exercise 13: If $x,y$ are complex, prove that
  $$ | | x | - | y | | \leq | x - y | $$

ghostofgarborg · Accepted Answer

By the triangle inequality
$$ |x| = | (x - y) + y| \leq |x-y| + |y| $$
so that
$$ |x| - |y| \leq |x-y| $$
Interchanging the roles of $x$ and $y$ in the above, we also have
$$ |y| - |x| \leq |x-y| $$
so that
$$ ||x| - |y|| \leq |x-y| $$

Amit Awasthi · Answer

\begin{proof}
      By Theorem $1.33(e)$ we know, for $z = y-x$,
      \begin{align*}
        |x + z| &\le |x| + |z|\
        |x + z| - |x| &\le |z|\
        |y| - |x| &\le |y-x|.
\end{align*}

Similarly, for $z = x-y$,
\begin{align*}
        |y + z| &\le |y| + |z|\
        |y + z| - |y| &\le |z|\
        |x| - |y| &\le |x-y|.
\end{align*}

Since $\max(|y| - |x|,|x| - |y|) = \big||x|-|y|\big|$,
\begin{align*}
        \big||x|-|y|\big| &\le |x-y|.
\end{align*}
\end{proof}

Exercise 1.13

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Exercise 1.13

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