Let
$$z_1 = a_1 + b_1i$$ $$z_2 = a_2 + b_2i$$
where
$$|z_j| = \sqrt{a_j^2 + b_j^2}$$
Prove
- $$|z_1 + z_2| \le |z_1| + |z_2|$$
- $$|z_1 + z_2| \ge |z_1| - |z_2|$$
- $$|z_1 - z_2| \ge |z_1| - |z_2|$$
- $$(2) \iff (3)$$
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
- $$LHS = \sqrt{(a_1+a_2)^2 + (b_1+b_2)^2}$$
$$RHS = \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2}$$
- $$LHS = \sqrt{(a_1+a_2)^2 + (b_1+b_2)^2}$$
$$RHS = \sqrt{a_1^2 + b_1^2} - \sqrt{a_2^2 + b_2^2}$$
- $$LHS = \sqrt{(a_1-a_2)^2 + (b_1-b_2)^2}$$
$$RHS = \sqrt{a_1^2 + b_1^2} - \sqrt{a_2^2 + b_2^2}$$
This looks to be precalculus, but I don't see it. Perhaps I got my complex analysis wrong?
- $$LHS_2 \ge LHS_3$$ so that proves $\implies$, but what about $\impliedby$?
Consider the conjugate $\bar{z} = x - yi$ of a complex number $z = x + yi$. It has some elementary properties such as:
Also, an obvious property of complex numbers is
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):
$|z| = |z + w - w| \leq |z + w| + |w| \implies |z| - |w| \leq |z + w|$