Prove that $\forall z \in \Bbb C : \lvert \Re(z) \rvert \le \lvert z \rvert \le \lvert \Re(z) \rvert + \lvert \Im(z)\rvert$ I'm having difficulties trying to prove these two complex inequalities :
$\forall z \in \Bbb C :$
$$\lvert \Re(z) \rvert \le \lvert z \rvert \le \lvert \Re(z) \rvert + \lvert \Im(z)\rvert$$
$$\lvert \Im(z) \rvert \le \lvert z \rvert \le \lvert \Re(z) \rvert + \lvert \Im(z) \rvert$$
By the defintion of the modulus of a complex number we have :
$$\lvert z \rvert = \lvert x+iy \rvert = \sqrt{x^2+y^2} = [(\Re(z))^2 + (\Im(z))^2]^{1/2}$$ 
If I square both side of the equality I get : $${\lvert z \rvert}^2 = (\Re(z))^2 + (\Im(z))^2$$ but I don't know how to continue from here. 

You have 
$$ |z| = \sqrt{(\Re z)^2 + (\Im z)^2} \geq \sqrt{(\Re z)^2} = |\Re z|$$
and
$$ |z| = \sqrt{(\Re z)^2 + (\Im z)^2} \leq \sqrt{|\Re z|^2 +2|\Re z| |\Im z|+ |\Im z|^2} = \sqrt{(|\Im z| + |\Re z|)^2} = |\Im z| + |\Re z|. $$
