Complex Analysis: Proofs on Re(z) & Im(z)

Question

How do you prove that

$-|z| \leq \Re(z) \leq |z|$ and $-|z| \leq \Im(z) \leq |z|$ ?

With $z=x+iy$ and $z \in \mathbb{C}$.

$\Re(z)= x$ and $\Im(z)=y$

I know that $|z| = \sqrt{x^2 +y^2}$ Similarly $- |z| = - \sqrt{x^2 +y^2}$

Answer 1

There's nothing much. It's an (almost) immediate observation which follows from the fact that a sum of two squares (of real numbers) is always bigger than (or just the same as) each one of the component squares; that is, $$x^2+y^2\ge x^2.$$ Interchanging the roles of $x$ and $y$ gives you a similar inequality. If you now take square roots of both sides, you get that $$\sqrt{x^2+y^2}\ge |x|,$$ which is equivalent to $$-\sqrt{x^2+y^2}\le x\le \sqrt{x^2+y^2},$$ as wanted.

The one for the imaginary part also follows, by performing $x\mapsto y.$