# Thread: Complex Analysis: Prove this Inequality

1. ## Complex Analysis: Prove this Inequality

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

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

3. 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.

4. 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 $\sqrt {\left| a \right|^2 + \left| b \right|^2 } \leqslant \sqrt {\left| a \right|^2 + 2\left| {ab} \right| + \left| b \right|^2 } = \left| a \right| + \left| b \right|$