$|Re (z)|\le|z|\qquad$ $(z := x + yi)$
$|Re (z)|= |x|=\sqrt{x^2}\qquad |z| = \sqrt{x^2 + y^2}$
$\sqrt{x^2}\le \sqrt{x^2 + y^2}$ (therefore $y^2 \ge 0$)
then the answer is $\sqrt{x²}\le \sqrt{x^2 + y^2}$ is this correct?
and do the same thing for $Im(z)$?
is this correct? Thanks!
On
Your argument is pretty much correct. Just to give a bit more intuition for what's going on, consider the diagram below, showing a complex number on the Argand diagram.
We see that $x=\text{Re }z, y = \text{Im }z$ are the two shorter sides of a right-angled triangle, and its hypothenuse is $\lvert z\rvert$. It is visually very clear that $\lvert z\rvert \geq \lvert x\rvert, \lvert y\rvert$.
It is almost correct. You should have added that, when you write $z$ as $x+yi$, your numbers $x$ and and $y$ are real (although, yes, this is usually implicit). And you also should have added that $\sqrt{x^2}=|x|=|\operatorname{Re}z|$.