Prove the triangle inequality involving complex numbers.

Question

Our eventual goal in this problem is to prove the triangle inequality involving complex numbers.

(a) Show that for every $z ∈ C$,

$|Re(z)| ≤ |z|$ and $|Im(z)| ≤ |z|$.

(b) Given $z$, $w ∈ C$, show that

$|z+w|^2 =|z|^2 +|w|^2 +2Re(zw')$.

(c) Using parts (a) and (b), prove the triangle inequality

$|z + w| ≤ |z| + |w|$.

This is what I got.

(a) By definition for a complex number $z = x + yi$,

$$|z|^2 = x^2 + y^2 = Re(z)^2 + Im(z)^2$$

From here,

$$|z|^2 ≥ Re(z)^2 \text{ and } |z|^2 ≥ Im(z)^2$$

And, finally,

$$|z| ≥ |Re(z)| \text{ and } |z| ≥ |Im(z)|$$

(b) $|z + w|^2 = (z + w)·(z + w)'$

$$= (z + w)·[z' + w']$$

$$= zz' + [zw' + z'w] + ww'$$

$$= |z|^2 + 2Re[zw'] + |w|^2$$

$$≤ |z|^2 + 2|zw'| + |w|^2$$

$$= |z|^2 + 2|z||w| + |w|^2$$

$$= (|z| + |w|)^2.$$

(c) Since both $|z+w|$ and $|z| + |w|$ are non-negative,

$$|z + w| ≤ |z| + |w|$$

Answer 1

Looks solid! Sadly, I'm only seeing it well after the OP, but to cut down on unanswered questions, here we go!

Answer 2

As I am just a student be precocious with my answer.

-Using basic complex number property.
On one side we have:
$|z+w|^2=(z+w)(z+w)^*= (z+w)(z^*+w^*)=zz^*+zw^*+wz^*+ww^*=|z|^2+|w|^2+2Re(zw*)$
On the other side we have:
$(|z|+|w|)^2=|z|^2+|w|^2+2|z||w|$

-Now as we know that $ \forall z=x+iy \in \mathbb{C} \Rightarrow Re(z)=x \leq |z|= \sqrt{x^2+y^2} \Rightarrow Re(zw^*)<|zw|$.

-Hence we can conclude that: $|z+w|^2 \leq (|z|+|w|)^2 \Rightarrow |z+w| \leq |z|+|w| $ Q.E.D.

Answer 3

Let $z,w$ be complex numbers,then we have that $||z|-|w|| \leq |z+w|$ and $||z|-|w|| \leq |z-w| $ furthermore $|z-w| \leq |z|+|w| $ also $ |z+w| \leq |z|+|w|$.

Proof:

Clearly $ |z|^{2}+|w|^{2} \leq |z|^{2}+2|w||y|+|w|^{2} $ Since we have that $2|z||w| \leq 0$ by definition of the modulo. So we have that $|z|^{2}+|w|^{2} \leq (|z+w|)^{2}$ which implies that $|z|+|w| \leq |z+w|$.

Now $|z|^{2}-|w|^{2} \leq |z|^{2}-2|w||y|+|w|^{2}$ .Since $2|z||w| \leq 0$ Again this leads use to conclude that $|z|^{2}-|w|^{2} \leq (|z-w|)^{2} \leq (|z+w|)^{2}$ since $|z|^{2}-2|w||y|-|w|^{2} \leq |z|^{2}+2|w||y|+|w|^{2}$ then we have that $|z-w| \leq |z|+|w|$.

Also $|z|=|z-w+w| \leq |z-w|+|w|$ (as shown above) $⇒|z|-|w|\leq|z-w|$, Similarly $|w|=|w-z+z|\leq|w-z|+|z| ⇒|w|-|z|\leq|w-z|$ multiplying both sides by -1 gives $-|z-w|\leq|z|-|w|$ so $-|z-w|\leq|z|-|w|\leq|z-w| ⇒||z|-|w||\leq|z-w|$.

$||z|-|-w||\leq|z-(-w)|⇒||z|-|w||\leq|z+w|$ since $|-w|=|w|$. And therefore $ ||z|-|w|| \leq |z-w|\leq|z|+|w|$ , $|z+w|\leq|z|+|w|$ has been shown.