We have an answer to "How can we prove that |z|≤Re(z)?"

Answered question

2022-01-17

How can we prove that

| z | \leq R e (z) ?

Answer & Explanation

nick1337

Expert2022-01-17Added 777 answers

Step 1
Trivial for negative real z, so we can make it stronger than that:
$| z | \leq | z | | \cos (a r g (z)) | = | R e (z) |$
Alternatively,
$| a + b i | \leq | a |$
$\Leftrightarrow | a + b i |^{2} \leq a^{2} \Leftrightarrow a^{2} + b^{2} \leq a^{2}$
and the last line is obvious.
It should also be clear geometrically; the hypotenuse is greater than the length of either leg of a right triangle.

star233

Skilled2022-01-17Added 403 answers

Step 1
Let
$z = x + i t$
Then
$R e (z) = x$
Now
$| z | = \sqrt{(x^{2} + y^{2})} \geq \sqrt{(x^{2})} = | x | \geq x = R e (z)$
Thus
$| z | \geq R e (z)$ (equality can hold off $x \geq 0$ and $y = 0$ )

alenahelenash

Expert2022-01-24Added 556 answers

Step 1 Let

z = x + i y

So,

R e (z) = x

and

(z) = \sqrt{x^{2} + y^{2}}

You would have to prove

\sqrt{x^{2} + y^{2}} \geq x

Notice that,

y^{2} \geq 0

. Then,

x^{2} + y^{2} \geq x^{2} \Rightarrow \sqrt{x^{2} + y^{2}} \geq | x | \geq x

where does it follow that,

| z | = \sqrt{x^{2} + y^{2}} \geq x = R e (z) .

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