I am reading Ahlfors' "Complex Analysis". Early in the book, he uses the fact that for $z \in \mathbb{C}$ we have
$$
-\lVert z\rVert \le \Re (z) \le \lVert z\rVert\qquad \text{and} \qquad -\lVert z\rVert \le \Im (z) \le \lVert z\rVert
$$
He says that these can inequalities can be derived from the definitions of the real and imaginary parts, as well as the definition of the absolute value of a complex number. These definitions are as follows:
$$
\Re (z) = \frac{z + \overline{z}}{2} \qquad \Im (z) = \frac{z -\overline{z}}{2i} \qquad \rVert z \rVert^2 = z \overline{z}
$$
I managed to prove the statement using the following method. I write out $z$ explicitly as $z = x + iy$ for some $x, y \in \mathbb{R}$. Using this I can show that these definitions are equivalent to
$$
\Re (z) = x \qquad \Im (z) = y \qquad \lVert z\rVert^2 = x^2 + y^2
$$
Using these new definition, that fact that $a^2 \ge 0\ \forall a \in \mathbb{R}$, and knowing that the real-valued function $f(x) = \sqrt{x}$ is monotonically increasing on $[0, \infty)$, I can show that
$$
\sqrt{x^2 + y^2} \ge \sqrt{x^2} = |x| \qquad \sqrt{x^2 + y^2} \ge \sqrt{y^2} = |y|
$$
which is equivalent to saying
$$
\lVert z\rVert \ge |\Re (z)| \qquad \lVert z\rVert \ge |\Im (z)|
$$
proving the statement.
I don't like the proof I got because I feel like it "backtracks" into doing grunt work. All the definitions given are written in such a way that you don't need to write out a complex number $z$ as $x + iy$, so I feel like going back to this is not a "clean" proof.
Up to this point, the book has proven previously that the absolute value of a complex number is distributive over addition and multiplication of complex numbers, that $\overline{\overline{z}} = z$, and the following properties (for $a,b \in \mathbb{C}$):
$$
\lVert a + b \rVert ^2 = \lVert a \rVert ^2 + \lVert b \rVert ^2 + 2 \Re\left(a \overline{b}\right) \qquad \quad \lVert a – b \rVert ^2 = \lVert a \rVert ^2 + \lVert b \rVert ^2 – 2 \Re\left(a \overline{b}\right)
$$
I tried using these properties to give a proof of the statement where I didn't have to write out $z = x+iy$ explicitly, but I didn't seem to be able to get anywhere. Does anyone know a way to prove this statement without backtracking as I did? Thank you!
Best Answer
Using the definitions $$ \Re (z) = \frac{z + \overline{z}}{2} \, , \, \Im (z) = \frac{z -\overline{z}}{2i} $$ you can compute $$ \bigl(\Re (z)\bigr)^2 + \bigl(\Im (z)\bigr)^2 = \left(\frac{z + \overline{z}}{2}\right)^2 + \left(\frac{z -\overline{z}}{2i} \right)^2 = z \overline{z} = \lVert z \rVert^2 $$ so that $$ \bigl(\Re (z)\bigr)^2\le \lVert z \rVert^2 \implies |\Re (z)| \le \lVert z \rVert $$ and similarly for the imaginary part.