Help with basic inequality (complex numbers)

Question

I want to prove

\begin{align} \lvert z\rvert &\geq \Re\{z\} \tag{1}\\ \lvert z\rvert &\geq \Im\{z\} \tag{2} \end{align}

I start with $z=x+iy$ so $$ \lvert z\rvert=\sqrt{x^2+y^2}\tag{3} $$

With the following (I guess it's valid for complex numbers?) $$ \lvert z\rvert =\sqrt{z^2} \iff \lvert z\rvert^2 =z^2 \tag{4} $$

I can write \begin{gather} \lvert z\rvert^2=x^2+y^2 \tag{5} \end{gather}

Using \begin{gather} \Re\{z\}=x \iff \Re\{z\}^2=x^2 \tag{6} \\ \Im\{z\}=y \iff \Im\{z\}^2=y^2 \tag{7} \end{gather} I can now write \begin{gather} \lvert z \rvert ^2=\Re\{z\}^2 +\Im\{z\}^2 \tag{8} \end{gather}

I'm stuck here.

What is the next step? Or should I stop here and conclude something from $(8)$?

Thanks!

Update: I not sure, but shouldn't we have absolute values in $(1)$ and $(2)$, i.e. $\lvert z\rvert \geq \lvert \Re\{z\}\rvert$ and $\lvert z\rvert \geq \lvert\Im\{z\}\rvert$?

Answer 1

Hint:

Since $\Im \{z\}$ is a real number, you know that $\Im \{z\}^2 \geq 0$.

Answer 2

Note that

$$\begin{cases}\lvert z\rvert=\sqrt{x^2+y^2}\ge x\\\\\lvert z\rvert=\sqrt{x^2+y^2}\ge y\end{cases}$$

are always true.

Indeed they are true when $x,y<0$ and for $x,y\ge0$ we can square and obtain

$$\begin{cases}x^2+y^2\ge x^2\\\\x^2+y^2\ge y^2\end{cases}$$

As an alternative, note that in polar form the equations are equivalent to

$$\begin{cases}\rho\ge \rho \cos \theta \iff \cos \theta \le1\\\\\rho\ge \rho \sin \theta \iff\sin \theta \le1\end{cases}$$