# Prove Basic Complex Number Inequalities

Let

$$z_1 = a_1 + b_1i$$ $$z_2 = a_2 + b_2i$$

where

$$|z_j| = \sqrt{a_j^2 + b_j^2}$$

Prove

1. $$|z_1 + z_2| \le |z_1| + |z_2|$$
2. $$|z_1 + z_2| \ge |z_1| - |z_2|$$
3. $$|z_1 - z_2| \ge |z_1| - |z_2|$$
4. $$(2) \iff (3)$$

I based the above on 3 and 4 in Schaum's Complex Variables below: I cannot use polar coordinates as those are presented later on. These are presented in the context of the absolute value function

1. $$LHS = \sqrt{(a_1+a_2)^2 + (b_1+b_2)^2}$$

$$RHS = \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2}$$

1. $$LHS = \sqrt{(a_1+a_2)^2 + (b_1+b_2)^2}$$

$$RHS = \sqrt{a_1^2 + b_1^2} - \sqrt{a_2^2 + b_2^2}$$

1. $$LHS = \sqrt{(a_1-a_2)^2 + (b_1-b_2)^2}$$

$$RHS = \sqrt{a_1^2 + b_1^2} - \sqrt{a_2^2 + b_2^2}$$

This looks to be precalculus, but I don't see it. Perhaps I got my complex analysis wrong?

1. $$LHS_2 \ge LHS_3$$ so that proves $\implies$, but what about $\impliedby$?
• Hint for 1,2,3: The Cauchy-Schwarz inequality yields $|u_1u_2 | + |v_1v_2| \le \sqrt{u_1^2 + v_1^2} \times \sqrt{u_2^2 + v_2^2}$ for all real $(u_1,u_2,v_1,v_2)$ Jul 12, 2016 at 22:51

Consider the conjugate $\bar{z} = x - yi$ of a complex number $z = x + yi$. It has some elementary properties such as:

• $\bar{\bar{z}} = z$
• $\overline{z \cdot w} = \bar{z} \cdot \bar{w}$
• $\overline{z + w} = \bar{z} + \bar{w}$
• $|z|^2=z \cdot \bar{z}$

Also, an obvious property of complex numbers is

• $x =Re(z) \leq |z| = \sqrt{x^2 + y^2}$

One verifies the above facts with a straightforward computation. Then the triangle inequality follows:

$|z + w|^2 = (z + w)\cdot \overline{(z + w)} = (z + w)\cdot(\bar{z} + \bar{w}) = |z|^2 + z\cdot\bar{w} + \bar{z} \cdot w + |w|^2$

But $\overline{z\cdot\bar{w}} = \bar{z} \cdot w$, and $\overline{z\cdot\bar{w}} + \bar{z} \cdot w = 2Re(\bar{z}\cdot w)$ - another straightforward computation. Also, $2Re(\bar{z}\cdot w) \leq 2 |\bar{z}\cdot w| = 2 |z||w|$ - (since $|z\cdot w|^2 = (z \cdot w)\cdot \overline{z \cdot w} = |z|^2 |w|^2$

Therfore,

$|z|^2 + z\cdot\bar{w} + \bar{z} \cdot w + |w|^2 = |z|^2 + 2Re(\bar{z}\cdot w) + |w|^2 \leq |z|^2 + 2|z||w| + |w|^2 =(|z| + |w|)^2$

Putting it all together

$|z + w|^2 \leq (|z| + |w|)^2 \implies |z + w| \leq (|z| + |w|)$

2 and 3 in your question follow by a neat trick using the triangle inequality, for (2):

$|z| = |z + w - w| \leq |z + w| + |w| \implies |z| - |w| \leq |z + w|$