Complex Analysis: Prove this Inequality

**joestevens** · September 6th 2010, 07:41 AM

Prove that for $z \in \mathbb{C}$ , $|z| \leq |Re \ z| + |Im \ z| \leq \sqrt{2}|z|$ .

**Plato** · September 6th 2010, 08:06 AM

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 $a=\text{Re}(z)~\&~ b=\text{Im}(z).$

**joestevens** · September 6th 2010, 08:22 AM

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.

**Plato** · September 6th 2010, 08:38 AM

Originally Posted by joestevens

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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