I am reading Ahlfors’ "Complex Analysis". Early in the book, he uses the fact that for $z in mathbb{C}$ we have

$$

-lVert zrVert le Re (z) le lVert zrVertqquad text{and} qquad -lVert zrVert le Im (z) le lVert zrVert

$$

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 zrVert^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 zrVert ge |Re (z)| qquad lVert zrVert 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 Releft(a overline{b}right) qquad quad lVert a – b rVert ^2 = lVert a rVert ^2 + lVert b rVert ^2 – 2 Releft(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!

Mathematics Asked on November 12, 2021

2 AnswersSo we need to prove $|Re(z)|leq |z|$ and $|Im(z)|leq |z|$. Notice $$boxed{rm Chatetusleq Hypothenus}$$

Answered by Aqua on November 12, 2021

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 $$ bigl(Re (z)bigr)^2le lVert z rVert^2 implies |Re (z)| le lVert z rVert $$ and similarly for the imaginary part.

Answered by Martin R on November 12, 2021

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