# Proving Complex Inequality $||z_1z_2|-1| \leq|z_1z_2 + i| \leq |z_1z_2| +1$.

The question at hand is that I have $z_1, z_2 \in \mathbb{C}$ two complex numbers.

I was able to prove the following:

$|z_1 z_2 + i|^2 - |z_1z_2|^2 = 1+ 2 \Im (z_1z_2)$

The second part of the question required to prove the following inequality using the previous equation:

$$||z_1z_2|-1| \leq|z_1z_2 + i| \leq |z_1z_2| +1$$

I managed to prove part of it using the triangle inequality:

$|z_1z_2 + i |\leq |z_1z_2| + |i| = |z_1z_2| +1$

i tried using the equation to prove the first part of the inequality, but i'm not getting anywhere.

Any help is really appreciated

• Hint: $|-i| = 1$, and the reverse triangle inequality. – Camille Feb 5 '17 at 10:45
• ah that works, thanks. but then I wouldn't have used the equation – user368063 Feb 5 '17 at 10:56

Substitute $w = z_1z_2$ for simplicity. Then you've proved

$$|w+i|^2 - |w|^2 = 1 + 2\Im w\tag{1}$$

and you want to prove

$$||w|-1|\leq |w+i|\leq |w|+1\tag{2}$$

These are all nonnegative real numbers, so $(2)$ is equivalent to

$$||w|-1|^2\leq |w+i|^2\leq (|w|+1)^2$$ and if we use $(1)$, $(2)$ is equivalent to

$$|w|^2-2|w|+1\leq |w|^2+2\Im w +1\leq |w|^2 +2|w|+1$$

Can you finish from here?

• if I let $w = x + iy$ then $|w| = \sqrt{x^2 + y^2}$ and $Im(w) = y$. Therefore we have $\sqrt{x^2 + y^2} > \sqrt{y^2} = |y| \geq y$. Thus $|w| \geq Im(w)$ and the inequality follows – user368063 Feb 5 '17 at 11:24
• does this work? – user368063 Feb 5 '17 at 11:25
• Yes, it works. Just note that you need to prove $|\operatorname{Im} w|\leq |w|$, not just $\operatorname{Im} w\leq |w|$. But, with your notation we have $|\operatorname{Im} w| = |y| =\sqrt{y^2}\leq \sqrt{x^2+y^2} = |w|$, @HexaFlexagon. – Ennar Feb 5 '17 at 11:29