0
$\begingroup$

I'm Really struggling on where to go with this proof which says the following:

Prove that, for $z \in \mathbb{C}$ $$\left|z\right| \le \left|\mathrm{Re}(z)\right| + \left|\mathrm{Im}(z)\right| \le \sqrt {2}\left|z\right|$$

Some examples to show that either inequality may be an equality as well would also help a lot.

  • 1
    Do you know the triangle inequality? – 2017-02-15
  • 1
    I'm trying to get to use it better but my lecturers keep saying it during explanations and I'm still unsure on how to use it efficiently. I do have this in my notes though, $||z_1| - |z_2|| \le |z_1 + z_2| \le |z_1|+|z_2|$ – 2017-02-15

2 Answers 2

2

We need to establish

$$\sqrt{a^2+b^2}\le|a|+|b|\le2\sqrt{a^2+b^2}$$

Now square each of them and validate

  • 0
    I've gotten something like: $|z|=\sqrt{x^2+y^2}\le\sqrt{x^2}+\sqrt{y^2}\le\sqrt{x^2+y^2}+\sqrt{x^2+y^2}$ but that gives me $2|z|$ – 2017-02-15
  • 0
    @M.Turner, But $$|z|=\sqrt{x^2+y^2}$$ – 2017-02-15
  • 0
    sorry, I'm still new to the mathjax. Now look at the edited comment and that's what I wanted to say. – 2017-02-15
  • 0
    @M.Turner, Now square each of them – 2017-02-15
  • 0
    That'll give me, $$x^2 + y^2 \le \left(\sqrt{x^2}+\sqrt{y^2}\right)^2 \le \left(\sqrt{x^2+y^2}+\sqrt{x^2+y^2}\right)^2.$$ Now what? This looks to be worse. – 2017-02-15
1

Note that triangle inequality gives $$|z|=|\text{Re}(z)+i\text{Im}(z)|\leq |\text{Re}(z)|+|i\text{Im}(z)|=|\text{Re}(z)|+|\text{Im}(z)|.$$ To see the second inequality, note that $$2|z|^2-(|\text{Re}(z)|+|\text{Im}(z)|)^2=|\text{Re}(z)|^2+|\text{Im}(z)|^2-2|\text{Re}(z)||\text{Im}(z)|=(|\text{Re}(z)|-|\text{Im}(z)|)^2\geq 0.$$ Thus $$2|z|^2\geq(|\text{Re}(z)|+|\text{Im}(z)|)^2,$$ take square roots on both sides yield the result.